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Instructions of Test-2 (STAT-101)
The due date of Test 2 is Monday, January 30, 2023, at 11:00 PM.
Test 2 covers the material of Weeks 5, 7 & 8
The test consists of 25 questions
10T/F (0.25 marks each) and 15MCQ (0.5marks each)
Total Marks = 10
You have only one attempt.
You have a time limit of 5 hours (300 minutes).
This assignment will be saved and submitted automatically when the time (5hrs) is expired.
This assignment can be saved and resumed at any point until the time (5hrs) has expired.
The time will continue to run if you leave the test.

STAT 101_SEU 00967775703091 Assignment _2_2023 2_نموذج

Review Test Submission: Assignment2-STAT101-
2022-23-2nd

User

Course (Current Semester – الفصل الحالي)STAT-101: Statistics *******************

Test Assignment2-STAT101-2022-23-2nd

Started 1/27/23 11:52 PM

Submitted 1/28/23 1:55 AM

Due Date 1/30/23 11:00 PM

Status Completed

Attempt
Score

9.75 out of 10 points

Time
Elapsed

2 hours, 3 minutes out of 5 hours

Instructions Instructions of Assignment-2(STAT-101)
The display date of Assignment 2 is Wednesday, January 25,

2023, 11:00 P.M.
The due date of Assignment 2 is Monday, January 30, 2023, at

11:00 PM.

Assignment 2 covers the material of Weeks 5, 7, & 8 (Chapters-6,

7 & 8)

The assignment consists of 25 questions
10T/F (0.25 marks each) and 15MCQ (0.5marks each)

Total Marks = 10

You have only one attempt.

You have a time limit of 5 hours (300 minutes).
This assignment will be saved and submitted automatically when

the time (5hrs) is expired.
This assignment can be saved and resumed at any point until the

time (5hrs) has expired.
The time will continue to run if you leave the test.

Good luck!

Saturday, January 28, 2023 1:56:01 AM AST

STAT 101_SEU 00967775703091 Assignment _2_2023 2_نموذج

Question 1

If the total area under standard normal probability distribution is k+1, then the value

of k is zero.

True

False

Question 2

If the z-score of normal distribution is –2.50, the mean of the distribution is 35 and the

standard deviation of normal distribution is 2, then the value of X for a normal

distribution is 40.

True

False

Question 3
Given that Z is a standard normal random variable. If P(Z > k)=0.0505, then the value of k is
1.64

True

False

Question 4

A confidence interval (or interval estimate) is a range (or an interval) of values used

to estimate the true value of a population parameter.

True

False

Question 5

The sample mean is not the best point estimate of the population mean.

True

False

Question 6

If the P-value for a one-sided test for testing a mean is 0.05, then the P-value for the

corresponding two-sided test would be 0.01.

True

False

Question 7

The probability of rejecting the null hypothesis when it is true is called Level of

significance.

True

False

Question 8

The alternative hypothesis for the following claim: “A car Company claims that its new car

will average more than 40 miles per gallon in the city” is H1: µ < 40.

True

False

STAT 101_SEU 00967775703091 Assignment _2_2023 2_نموذج

Question 9

The alternative hypothesis for the following claim: “A motorbike company claims that its new

model will give an average at least 60 km/l on a long route” is Ha: µ < 60.

True

False

Question 10

If the original claim says that the mean working hours in a day are same for men and

women in a company. Then symbolically it is represented as p1 = p2.

True

False

Question 11

You are given the following hypothesis test:

H0: μ=100

H1: μ ≠ 100
The calculated test statistic z = –1.0, and the critical value of z = ±1.97. Then, the

decision would be to:

Reject H0 since z < –1.97

Reject H0 since –1.97 < z < 1.97

Fail to reject H0 since –1.97 < z < 1.97

Fail to reject H0 since z < –1.97

Question 12

A prescription allergy medicine is supposed to contain an average of 245 parts per

million (ppm) of active ingredient. The manufacturer periodically collects data to

determine if the production process is working properly. A random sample of 64 pills

has a mean of 250 ppm with a standard deviation of 12

ppm.

Let µ denotes the average amount of the active ingredient in pills of this allergy

medicine. The null and alternative hypotheses are as H0: µ = 245, Ha:µ ≠ 245. The

level of significance is 1%.

The t-test statistic is 3.33 with a P-value of 0.0014. What is the correct conclusion?

The mean amount of active ingredient in pills of this allergy medicine is equal to 245

ppm.

The mean amount of active ingredient in pills of this allergy medicine is equal to 250

ppm.

The mean amount of active ingredient in pills of this allergy medicine is not equal to 245

ppm.

The mean amount of active ingredient in pills of this allergy medicine is greater than 245

ppm.

Question 13

Among 169 Egyptian-African men, the mean systolic blood pressure was 145 mmHg

with a standard deviation of 26. The t-test statistic to conclude that the mean systolic

blood pressure for a population of Egyptian-African men is greater than 142 is

-2.5

-1.3

1.5

-1.5

STAT 101_SEU 00967775703091 Assignment _2_2023 2_نموذج

Question 14

The degree of confidence is equal to:

1-α

1-β

Question 15

When carrying out a large sample test of H0: µ0 = 50, Ha: µ0 < 50, we reject H0 at

level of significance α when the calculated test statistic is:

Greater than zα

Less than – zα

Greater than zα/2

Less than zα

Question 16

A sample of 100 body temperatures has a mean of 98.6 oF. Assume that σ is known to

be 0.5 oF. Use a 0.05 significance level to test the claim that the mean body

temperature of the population is equal to 98.5 oF, as is commonly believed. What is

the value of test statistic for this testing?

1.0

3.0

-2.0

2.0

Question 17

With H0: μ = 100, Ha: μ < 100, the test statistic is z = – 1.75. Using a 0.05

significance level, the P-value and the conclusion about null hypothesis are (Given

that P(z < 1.75) =0.9599)

0.0401; reject H0

0.9599; fail to reject H0

0.0401; fail to reject H0

0.9599; reject H0

Question 18

A passing student is failed by an examiner, it is an example of:

Type-I error

Type-II error

Best Decision

All of above

Question 19

The confidence interval, 0.548 < p < 0.834 is obtained for a population proportion, p.

The margin of error, E using these confidence interval limits is

0.143

0.286

0.691

1.382

STAT 101_SEU 00967775703091 Assignment _2_2023 2_نموذج

Question 20

If the point estimate 𝑝 ̂ is 0.8 and the lower confidence limit is 0.6, then the upper

confidence limit is:

1.0

0.7

0.6

0.4

Question 21

If the Margin of error E is 0.5 and the upper confidence limit is 9, then the lower

confidence limit is:

Question 22
Evaluate P(-1< Z< 2), where P(Z < 2)=0.9772 and P(Z< -1)=0.1587

c. -0.1359

d. 0.8185

b. 0.1359

a. -0.8185

Question 23

The normal probability distribution curve is symmetrical about mean µ. Then P(X <

μ) = P(X > μ) is equal to

0.25

0.50

0.75

STAT 101_SEU 00967775703091 Assignment _2_2023 2_نموذج

Question 24
Which of the following is NOT true regarding the normal distribution?

d. The points of the curve meet the X-axis at z = –3 and z = 3

b. It has a single peak

c. It is symmetrical

a. Mean, median and mode are all equal

Question 25
Assume that the thermometer readings are normally distributed with a mean of 0°C and a
standard deviation of 1°C for freezing water. If one thermometer is randomly selected, find the
probability that it reads (at the freezing point of water) greater than -1.75 degrees.

a. 0.0401

b. -0.9599

c. 0.9599

d. None

Lecture Slides

Elementary Statistics
Eleventh Edition

and the Triola Statistics Series

by Mario F. Triola

Chapter 7

Estimates and Sample Sizes

7-1

Review and Preview

7-2 Estimating a Population

Proportion

7-3 Estimating a Population Mean: σ Know

7-4 Estimating a Population Mean: σ Not Known

7-5 Estimating a Population Variance

Section 7-1

Review and Preview

Review

❖ Chapters 2 & 3 we used “descriptive
statistics” when we summarized data using
tools such as graphs, and statistics such as
the mean and standard deviation.

❖ Chapter 6 we introduced critical values:
z denotes the z score with an area of  to
its right.
If 

0.025, the critical value is z0.025 = 1.96.
That is, the critical value z0.025 = 1.96 has an
area of 0.025 to its right.

Preview

❖ The two major activities of inferential
statistics are (1) to use sample data to
estimate values of a population parameters,
and (2) to test hypotheses or claims made
about population parameters.

❖ We introduce methods for estimating values
of these important population parameters:
proportions, means, and variances.

❖ We also present methods for determining
sample sizes necessary to estimate those
parameters.

This chapter presents the beginning of
inferential statistics.

Section 7-2

Estimating a Population

Proportion

Key Concept

In this section we present methods for using a

sample proportion to estimate the value of a

population proportion.

• The sample proportion is the best point

estimate of the population proportion.

• We can use a sample proportion to construct a

confidence interval to estimate the true value

of a population proportion, and we should

know how to interpret such confidence

intervals.

• We should know how to find the sample size

necessary to estimate a population proportion.

Definition

A point estimate is a single value (or

point) used to approximate a population

parameter.

The sample proportion p is the best

point estimate of the population

proportion p.

Definition

Example:

Because the sample proportion is the best
point estimate of the population proportion, we
conclude that the best point estimate of p is
0.70. When using the sample results to
estimate the percentage of all adults in the
United States who believe in global warming,
the best estimate is 70%.

In the Chapter Problem we noted that in a Pew
Research Center poll, 70% of 1501 randomly
selected adults in the United States believe in
global warming, so the sample proportion is

= 0.70. Find the best point estimate of the
proportion of all adults in the United States
who believe in global warming.

p̂

Definition

A confidence interval (or interval

estimate) is a range (or an interval)

of values used to estimate the true

value of a population parameter. A

confidence interval is sometimes

abbreviated as CI.

A confidence level is the probability 1 –  (often

expressed as the equivalent percentage value)

that the confidence interval actually does contain

the population parameter, assuming that the

estimation process is repeated a large number of

times. (The confidence level is also called degree

of confidence, or the confidence coefficient.)

Most common choices are 90%, 95%, or 99%.

( = 10%), (

 = 5%

), ( = 1%)

Definition

We must be careful to interpret confidence intervals
correctly. There is a correct interpretation and many
different and creative incorrect interpretations of the
confidence interval 0.677 < p < 0.723.

“We are 95% confident that the interval from 0.677 to
0.723 actually does contain the true value of the
population proportion p.”

This means that if we were to select many different
samples of size 1501 and construct the corresponding
confidence intervals, 95% of them would actually
contain the value of the

population proportion p.

(Note that in this correct interpretation, the level of
95% refers to the success rate of the process being
used to estimate the proportion.)

Interpreting a Confidence Interval

Know the correct interpretation of a
confidence interval.

Caution

Confidence intervals can be used
informally to compare different data
sets, but the overlapping of
confidence intervals should not be
used for making formal and final
conclusions about equality of
proportions.

Caution

Critical Values

A standard z score can be used to distinguish
between sample statistics that are likely to
occur and those

that are unlikely to occur.

Such
a z score is called a critical value. Critical values
are based on the following observations:

1. Under certain conditions, the sampling
distribution of sample proportions can be
approximated by a normal

distribution.

Critical Values

2. A z score associated with a sample
proportion has a probability of /2 of falling in
the right tail.

Critical Values

3. The z score separating the right-tail region is

commonly denoted by z/2 and is referred to

as a critical value because it is on the

borderline separating z scores from sample

proportions that are likely to occur from those

that are unlikely to occur.

Definition

A critical value is the number on the

borderline separating sample statistics

that are likely to occur from those that are

unlikely to occur. The number z/2 is a

critical value that is a z score with the

property that it separates an area of /2 in

the

right tail of the standard normal

distribution.

The Critical Value z2

Notation for Critical Value

The critical value z/2 is the positive z value

that is at the vertical boundary separating an

area of /2 in the right tail of the standard

normal distribution. (The value of –z/2 is at

the vertical boundary for the area of /2 in the

left tail.) The subscript /2 is simply a

reminder that the z score separates an area of

/2 in the right tail of the standard normal

distribution.

Finding z2 for a 95%

Confidence Level

-z

z2

Critical Values

 2 = 2.5% = .025

 = 5%

z2 = 1.96−+

Use Table A-2 to find a z score of 1.96

 = 0.05

Finding z2 for a 95%

Confidence Level – cont

Definition

When data from a simple random sample are

used to estimate a population proportion p, the

margin of error, denoted by E, is the maximum

likely difference (with probability 1 – , such as

0.95) between the observed proportion and

the true value of the population proportion p.

The margin of error E is also called the

maximum error of the estimate and can be found

by multiplying the critical value and the standard

deviation of the sample proportions:

p̂

Margin of Error for

Proportions

ˆ ˆ

pq
E

n
=

p = population proportion

Confidence Interval for Estimating

a Population

Proportion p

= sample proportion

n = number of sample values

E = margin of error

z/2 = z score separating an area of /2 in
the right tail of the standard normal

distribution

p̂

Confidence Interval for Estimating

a Population Proportion p

1. The sample is a simple random sample.

2. The conditions for the binomial distribution

are satisfied: there is a fixed number of

trials, the trials are independent, there are

two categories of outcomes, and the

probabilities remain constant for each trial.

3. There are at least 5 successes and 5

failures.

Confidence Interval for Estimating

a Population Proportion p

p – E < < + Eˆ p̂p

where

ˆ ˆpq
E z

n


p – E < < +

p + E

ˆ ˆ

Confidence Interval for Estimating

a Population Proportion p

(p – E, p + E)ˆ ˆ

Round-Off Rule for

Confidence Interval Estimates of p

Round the confidence interval limits

for p to

three significant digits.

1. Verify that the required assumptions are

satisfied.

(The sample is a simple random sample, the

conditions for the binomial distribution are satisfied,

and the normal distribution can be used to

approximate the distribution of sample proportions

because np  5, and nq  5 are both satisfied.)

2. Refer to Table A-2 and find the critical value z /2 that

corresponds to the desired confidence level.

3. Evaluate the margin of error

Procedure for Constructing

a Confidence Interval for p

2
ˆ ˆE z pq n=

4. Using the value of the calculated margin of error, E

and the value of the sample proportion, p, find the

values of p – E and p + E. Substitute those values

in the general format for the confidence interval:

ˆ
ˆ ˆ

p – E < p < p + Eˆ ˆ

5. Round the resulting confidence interval limits to

three significant digits.

Procedure for Constructing

a Confidence Interval for p – cont

Example:

a. Find the margin of error E that corresponds to a

95% confidence level.

b. Find the 95% confidence interval estimate of the

population proportion p.

c. Based on the results, can we safely conclude that

the majority of adults believe in global warming?

d. Assuming that you are a newspaper reporter, write

a brief statement that accurately describes the

results and includes all of the relevant information.

In the Chapter Problem we noted that a Pew Research

Center poll of 1501 randomly selected U.S. adults

showed that 70% of the respondents believe in global

warming. The sample results are n = 1501, and ˆ 0.70p =

Requirement check: simple random sample;
fixed number of trials, 1501; trials are
independent; two categories of outcomes
(believes or does not); probability remains
constant. Note: number of successes and
failures are both at least 5.

Example:

a) Use the formula to find the margin of error.

( )

0 70 0 30
1 96

1501

0 02318

ˆ ˆ . .
.

pq
E z



= =

b) The 95% confidence interval:

Example:

ˆ ˆp E p p E−   +

0.70 − 0.023183  p  0.70 + 0.023183

0.677  p  0.723

c) Based on the confidence interval

obtained in part (b), it does appear that

the proportion of adults who believe in

global warming is greater than 0.5 (or

50%), so we can safely conclude that the

majority of adults believe in global

warming. Because the limits of 0.677 and

0.723 are likely to contain the true

population proportion, it appears that the

population proportion is a value greater

than 0.5.

Example:

d) Here is one statement that summarizes

the results: 70% of United States adults

believe that the earth is getting warmer.

That percentage is based on a Pew

Research Center poll of 1501 randomly

selected adults in the United States. In

theory, in 95% of such polls, the

percentage should differ by no more than

2.3 percentage points in either direction

from the percentage that would be found

by interviewing all adults in the United

States.

Example:

Analyzing Polls

When analyzing polls consider:

1. The sample should be a simple random sample,

not an inappropriate sample (such as a voluntary

response sample).

2. The confidence level should be provided. (It is

often 95%, but media reports often neglect to

identify it.)

3. The sample size should be provided. (It is usually

provided by the media, but not always.)

4. Except for relatively rare cases, the quality of the

poll results depends on the sampling method and

the size of the sample, but the size of the

population is usually not a factor.

Caution

Never follow the common misconception

that poll results are unreliable if the

sample size is a small percentage of the

population size. The population size is

usually not a factor in determining the

reliability of a poll.

Sample Size

Suppose we want to collect sample

data in order to estimate some

population proportion. The question is

how many sample items must be

obtained?

Determining Sample Size

(solve for n by algebra)

( )2 ˆp q  2Z n = ˆ
E 2

  2zE =
p qˆ ˆ
n

Sample Size for Estimating

Proportion p

When an estimate of p is known: ˆ

ˆ( )2 p qn = ˆ
E 2

  2z

When no estimate of p is known:

( )2 0.25n =
E 2

  2 z
ˆ

Round-Off Rule for Determining

Sample Size

If the computed sample size n is not

a whole number, round the value of n

to the next larger whole number.

Example:

The Internet is affecting us all in many

different ways, so there are many reasons for

estimating the proportion of adults who use

it. Assume that a manager for E-Bay wants to

determine the current percentage of U.S.

adults who now use the Internet. How many

adults must be surveyed in order to be 95%

confident that the sample percentage is in

error by no more than three percentage

points?

a. In 2006, 73% of adults used the Internet.

b. No known possible value of the proportion.

a) Use

To be 95% confident
that our sample
percentage is within
three percentage
points of the true
percentage for all
adults, we should
obtain a simple
random sample of 842
adults.

Example:

ˆ ˆ ˆ0.73 and 1 0.27

0.05 so 1.96

0.03

p q p



= = − =

= =

( )

( ) ( )( )

( )

ˆ ˆ

1.96 0.73 0.27

0.03

841.3104

842

z pq
n


=

b) Use

To be 95% confident that
our sample percentage
is within three
percentage points of the
true percentage for all
adults, we should obtain
a simple random sample
of 1068 adults.

Example:

 = 0.05 so z
 2

= 1.96

E = 0.03

( )

0.25

1.96 0.25

0.03

1067.1111

1068

z
n


=

Finding the Point Estimate and E

from a Confidence Interval

Margin of Error:

E = (upper confidence limit) — (lower confidence limit)

Point estimate of p:

p =
(upper confidence limit) + (lower confidence limit)

2
ˆ

Recap

In this section we have discussed:

❖ Point estimates.

❖ Confidence intervals.

❖ Confidence levels.

❖ Critical values.

❖ Margin of error.

❖ Determining sample sizes.

Section 7-3

Estimating a Population

Mean:  Known

Key Concept

This section presents methods for

estimating a population mean. In

addition to knowing the values of the

sample data or statistics, we must also

know the value of the population

standard deviation, .

Here are three key concepts that

should be learned in this section:

Key Concept

1. We should know that the sample mean

is the best point estimate of the

population mean .

2. We should learn how to use sample data

to construct a confidence interval for

estimating the value of a population mean,

and we should know how to interpret such

confidence intervals.

3. We should develop the ability to determine

the sample size necessary to estimate a

population mean.

Point Estimate of the

Population Mean

The sample mean x is the best point estimate

of the population mean µ.

Confidence Interval for

Estimating a Population Mean

(with  Known)

 = population mean

 = population standard deviation

= sample mean

n = number of sample values

E = margin of error

z/2 = z score separating an area of /2 in the

right tail of the standard normal

distribution

Confidence Interval for

Estimating a Population Mean

(with  Known)

1. The sample is a simple random sample.

(All samples of the same size have an

equal chance of being selected.)

2. The value of the population standard

deviation  is known.

3. Either or both of these conditions is

satisfied: The population is normally

distributed or n > 30.

Confidence Interval for

Estimating a Population Mean

(with  Known)

x − E    x + E where E = z

 2




or x  E

or x − E,x + E( )

Definition

The two values x – E and x + E are

called confidence interval limits.

1. For all populations, the sample mean x is an unbiased
estimator of the population mean , meaning that the
distribution of sample means tends to center about
the value of the population mean .

2. For many populations, the distribution of sample
means x tends to be more consistent (with less
variation) than the distributions of other sample
statistics.

Sample Mean

Procedure for Constructing a

Confidence Interval for µ (with Known )

1. Verify that the requirements are satisfied.

2. Refer to Table A-2 or use technology to find the
critical value z2 that corresponds to the desired
confidence level.

3. Evaluate the margin of error

5. Round using the confidence intervals round-off

rules.

4. Find the values of Substitute

those values in the general format of the
confidence interval:

2E z n =

x − E and x + E.

x − E    x + E

1. When using the original set of data, round the

confidence interval limits to one more decimal

place than used in original set of data.

2. When the original set of data is unknown and only

the summary statistics (n, x, s) are used, round the

confidence interval limits to the same number of

decimal places used for the sample mean.

Round-Off Rule for Confidence

Intervals Used to Estimate µ

Example:

People have died in boat and aircraft accidents
because an obsolete estimate of the mean
weight of men was used. In recent decades, the
mean weight of men has increased
considerably, so we need to update our
estimate of that mean so that boats, aircraft,
elevators, and other such devices do not
become dangerously overloaded. Using the
weights of men from Data Set 1 in Appendix B,
we obtain these sample statistics for the simple
random sample: n = 40 and = 172.55 lb.
Research from several other sources suggests
that the population of weights of men has a
standard deviation given by  = 26 lb.

Example:

a. Find the best point estimate of the mean
weight of the population of all men.

b. Construct a 95% confidence interval
estimate of the mean weight of all men.

c. What do the results suggest about the mean
weight of 166.3 lb that was used to
determine the safe passenger capacity of
water vessels in 1960 (as given in the
National Transportation and Safety Board
safety recommendation M-04-04)?

Example:

a. The sample mean of 172.55 lb is the best point
estimate of the mean weight of the population of all
men.

X – E <  < x + E

b. A 95% confidence interval or 0.95 implies
= 0.05, so z/2 = 1.96.
Calculate the margin of error.

Construct the confidence interval.

x − E    x − E

172.55 − 8.0574835    172.55 + 8.0574835

164.49    180.61

E = z/2
n

σ
=1.96

26
=8.0574835

Example:

c. Based on the confidence interval, it is
possible that the mean weight of 166.3 lb
used in 1960 could be the mean weight of
men today. However, the best point estimate
of 172.55 lb suggests that the mean weight
of men is now considerably greater than
166.3 lb. Considering that an underestimate
of the mean weight of men could result in
lives lost through overloaded boats and
aircraft, these results strongly suggest that
additional data should be collected.
(Additional data have been collected, and the
assumed mean weight of men has been
increased.)

Finding a Sample Size for

Estimating a Population Mean

(z/2) • 
n =

 = population mean

σ = population standard deviation

= sample mean

E = desired margin of error

zα/2 = z score separating an area of /2 in the right tail of

the standard normal distribution

Round-Off Rule for Sample Size n

If the computed sample size n is not a

whole number, round the value of n up

to the next larger whole number.

Finding the Sample Size n
When  is Unknown

1. Use the range rule of thumb (see Section 3-3)
to estimate the standard deviation as
follows:   range/4.

2. Start the sample collection process without
knowing  and, using the first several values,
calculate the sample standard deviation s and
use it in place of  . The estimated value of 
can then be improved as more sample data
are obtained, and the sample size can be
refined accordingly.

3. Estimate the value of  by using the results
of some other study that was done earlier.

Example:

 = 0.05

 /2 = 0.025

z / 2 = 1.96

E = 3

 = 15

n = 1.96 • 15 = 96.04 = 97

With a simple random sample of only

97 statistics students, we will be 95%

confident that the sample mean is

within 3 IQ points of the true

population mean .

Assume that we want to estimate the mean IQ score for

the population of statistics students. How many

statistics students must be randomly selected for IQ

tests if we want 95% confidence that the sample mean

is within 3 IQ points of the population mean, and

population standard deviation is 15 ?

Recap

In this section we have discussed:

❖ Margin of error.

❖ Confidence interval estimate of the

population mean with σ known.

❖ Round off rules.

❖ Sample size for estimating the mean μ.

Section 7-4

Estimating a Population

Mean:  Not Known

Key Concept

This section presents methods for estimating

a population mean when the population

standard deviation is not known. With σ

unknown, we use the Student t distribution

assuming that the relevant requirements are

satisfied.

The sample mean is the best point

estimate of the population mean.

Sample Mean

If the distribution of a population is essentially

normal, then the distribution of

is a

Student t Distribution

for all samples of
size n. It is often referred to as a t distribution
and is used to find critical values denoted by
t/2.

t =
x – µ

s
n

Student t Distribution

degrees of freedom = n – 1

in this section.

Definition

The number of degrees of freedom for a

collection of sample data is the number of

sample values that can vary after certain

restrictions have been imposed on all data

values. The degree of freedom is often

abbreviated df.

Margin of Error E for Estimate of 

(With σ Not Known)

Formula 7-6

where t2 has n – 1 degrees of freedom.

n
s

E = t 2

Table A-3 lists values for tα/2

 = population mean

= sample mean

s = sample standard deviation

n = number of sample values

E = margin of error

t/2 = critical t value separating an area of /2

in the right tail of the t distribution

Notation

where E = t/2
n
s

x – E < µ < x + E

t/2 found in Table A-3

Confidence Interval for the

Estimate of μ (With σ Not Known)

df = n – 1

2. Using n – 1 degrees of freedom, refer to Table A-3 or use
technology to find the critical value t2 that corresponds to
the desired confidence level.

Procedure for Constructing a

Confidence Interval for µ

(With σ Unknown)

1. Verify that the requirements are satisfied.

3. Evaluate the margin of error E = t2 • s / n .

4. Find the values of Substitute those
values in the general format for the confidence interval:

5. Round the resulting confidence interval limits.

x − E and x + E.

x − E    x + E

Example:

A common claim is that garlic lowers cholesterol

levels. In a test of the effectiveness of garlic, 49

subjects were treated with doses of raw garlic, and

their cholesterol levels were measured before and

after the treatment. The changes in their levels of LDL

cholesterol (in mg/dL) have a mean of 0.4 and a

standard deviation of 21.0. Use the sample statistics of

n = 49, = 0.4 and s = 21.0 to construct a 95%

confidence interval estimate of the mean net change in

LDL cholesterol after the garlic treatment. What does

the confidence interval suggest about the

effectiveness of garlic in reducing LDL cholesterol?

Example:

Requirements are satisfied: simple random

sample and n = 49 (i.e., n > 30).

21 0
2 009 6 027

.
. .E t

n



= = =

95% implies α = 0.05.

With n = 49, the df = 49 – 1 = 48

Closest df is 50, two tails, so t/2 = 2.009

Using t/2 = 2.009, s = 21.0 and n = 49 the

margin of error is:

Example:

Construct the confidence

interval:
x = 0.4, E = 6.027

We are 95% confident that the limits of –5.6 and 6.4

actually do contain the value of , the mean of the

changes in LDL cholesterol for the population. Because

the confidence interval limits contain the value of 0, it is

very possible that the mean of the changes in LDL

cholesterol is equal to 0, suggesting that the garlic

treatment did not affect the LDL cholesterol levels. It

does not appear that the garlic treatment is effective in

lowering LDL cholesterol.

x − E    x + E

0.4 − 6.027    0.4 + 6.027

−5.6    6.4

Important Properties of the

Student t Distribution
1. The Student t distribution is different for different sample sizes

(see the following slide, for the cases n = 3 and n = 12).

2. The Student t distribution has the same general symmetric bell
shape as the standard normal distribution but it reflects the
greater variability (with wider distributions) that is expected
with small samples.

3. The Student t distribution has a mean of t = 0 (just as the
standard normal distribution has a mean of z = 0).

4. The standard deviation of the Student t distribution varies with
the sample size and is greater than 1 (unlike the standard
normal distribution, which has a  = 1).

5. As the sample size n gets larger, the Student t distribution gets
closer to the normal distribution.

Student t Distributions for

n = 3 and n = 12

Figure 7-5

Choosing the Appropriate Distribution

Use the normal (z)
distribution

 known and normally
distributed population
or
 known and n > 30

Use t distribution  not known and
normally distributed
population
or
 not known and n > 30

Use a nonparametric
method or
bootstrapping

Population is not
normally distributed
and n ≤ 30

Point estimate of µ:

x = (upper confidence limit) + (lower confidence limit)

Margin of Error:

E = (upper confidence limit) – (lower confidence limit)

Finding the Point Estimate

and E from a Confidence Interval

Confidence Intervals for

Comparing Data

As in Sections 7-2 and 7-3, confidence

intervals can be used informally to

compare different data sets, but the

overlapping of confidence intervals should

not be used for making formal and final

conclusions about equality of means.

Recap

In this section we have discussed:

❖ Student t distribution.

❖ Degrees of freedom.

❖ Margin of error.

❖ Confidence intervals for μ with σ unknown.

❖ Choosing the appropriate distribution.

❖ Point estimates.

❖ Using confidence intervals to compare data.

Lecture Slides

Elementary Statistic

Eleventh Edition

and the Triola Statistics Series

by Mario F. Triola

Chapter 8

Hypothesis

Test

ing

8-1

Review and Preview

8-2 Basics of

Hypothesis Testing

8-3 Testing a Claim about a Proportio

8-4 Testing a Claim About a Mean: σ Known

8-5 Testing a Claim About a Mean: σ Not Known

Section 8-1

Review and Preview

Review
In Chapters 2 and 3 we used “descriptive statistics”
when we summarized data using tools such as graphs,
and statistics such as the mean and standard
deviation.

Methods of inferential statistics use sample data to
make an inference or conclusion about a population.
The two main activities of inferential statistics are
using sample data to (1) estimate a population
parameter (such as estimating a population parameter
with a confidence interval), and (2) test a hypothesis or
claim about a population parameter.

In Chapter 7 we presented methods for estimating a
population parameter with a confidence interval, and in
this chapter we present the method of hypothesis
testing.

Definitions

In statistics, a hypothesis is a claim or
statement about a property of a population.

A hypothesis test (or test of significance) is a
standard procedure for testing a claim about a
property of a population.

Main Objective

The main objective of this chapter is to

develop the ability to conduct hypothesis

tests for claims made about a:

– population proportion p,

– a population mean ,

– or a population standard deviation .

Examples of Hypotheses that can be Tested

• Genetics: The Genetics & IVF Institute claims
that its XSORT method allows couples to increase
the probability of having a baby girl.

• Business: A newspaper headline makes the
claim that most workers get their jobs through
networking.

• Medicine: Medical researchers claim that when
people with colds are treated with echinacea, the
treatment has no effect.

Examples of Hypotheses that can be Tested

• Aircraft Safety: The Federal Aviation
Administration claims that the mean weight of an
airline passenger (including carry-on baggage) is
greater than 185 lb, which it was 20 years ago.

• Quality Control: When new equipment is used
to manufacture aircraft altimeters, the new
altimeters are better because the variation in the
errors is reduced so that the readings are more
consistent. (In many industries, the quality of
goods and services can often be improved by
reducing variation.)

Caution

When conducting hypothesis tests as

described in this chapter and the

following chapters, instead of jumping

directly to procedures and calculations,

be sure to consider the context of the

data, the source of the data, and the

sampling method used to obtain the

sample data.

Section 8-2

Basics of Hypothesis

Testing

Key Concept

This section presents individual

components of a hypothesis test. We should

know and understand the following:

• How to identify the null hypothesis and alternative
hypothesis from a given claim, and how to express
both in symbolic form

• How to calculate the value of the test statistic, given a
claim and sample

data

• How to identify the critical value(s), given a
significance level

• How to identify the P-value, given a value of the test
statistic

• How to state the conclusion about a claim in simple
and nontechnical terms

Part 1:

The Basics of Hypothesis Testing

Rare Event Rule for

Inferential Statistics

If, under a given assumption, the

probability of a particular observed event

is exceptionally small, we conclude that

the assumption is probably not correct.

Components of a

Formal Hypothesis

Test

Null Hypothesis:

• The null hypothesis (denoted by H0) is
a statement that the value of a
population parameter (such as
proportion, mean, or standard
deviation) is equal to some claimed
value.

• We test the null hypothesis directly.

• Either reject H0 or fail to reject H0.

Alternative Hypothesis:

• The alternative hypothesis (denoted
by H1 or Ha or HA) is the statement that
the parameter has a value that
somehow differs from the null

hypothesis.

• The symbolic form of the alternative
hypothesis must use one of these
symbols: , <, >.

Note about Forming Your

Own Claims (Hypotheses)

If you are conducting a study and want
to use a hypothesis test to support
your claim, the claim must be worded
so that it becomes the alternative
hypothesis.

Note about Identifying

H0 and H1

Figure 8-2

Example:

Consider the claim that the mean weight of

airline passengers (including carry-on

baggage) is at most 195 lb (the current value

used by the Federal Aviation Administration).

Follow the three-step procedure outlined in

Figure 8-2 to identify the null hypothesis and

the alternative hypothesis.

Example:

Step 1: Express the given claim in symbolic

form. The claim that the mean is at

most 195 lb is expressed in symbolic

form as  ≤ 195 lb.

Step 2: If  ≤ 195 lb is false, then  > 195 lb

must be true.

Example:

Step 3: Of the two symbolic expressions

 ≤ 195 lb and  > 195 lb,

we see that  > 195 lb does not contain

equality, so we let the alternative

hypothesis H1 be  > 195 lb.

Also, the null hypothesis must be a

statement that the mean equals 195 lb, so

we let H0 be  = 195 lb.

Note that the

original claim

that the mean is at most

195 lb is neither the alternative hypothesis nor the null

hypothesis.

(However, we would be able to address the original claim

upon completion of a hypothesis test.)

The test statistic is a value used in making

a decision about the null hypothesis, and is

found by converting the sample statistic to

a score with the assumption that the null

hypothesis is true.

Test Statistic

Test Statistic – Formulas

Test statistic for

proportion

p̂ − p

Test statistic

for mean

z =
x − 



t =
x − 

Example:

Let’s again consider the claim that the XSORT

method of gender selection increases the

likelihood of having a baby girl.

Preliminary results from a test of the XSORT

method of gender selection involved 14 couples

who gave birth to 13 girls and 1 boy. Use the given

claim and the preliminary results to calculate the

value of the test statistic.

Use the format of the test statistic given above, so

that a normal distribution is used to approximate a

binomial distribution.

(There are other exact methods that do not use the

normal approximation.)

Example:

The claim that the XSORT method of gender

selection increases the likelihood of having a

baby girl results in the following null and

alternative hypotheses H0: p = 0.5 and

H1: p > 0.5.

We work under the assumption that the null

hypothesis is true with p = 0.5.

The sample proportion of 13 girls in 14 births

results in . Using p = 0.5,

and n = 14,

we find the value of the test statistic as follows:

p̂ = 13 14 = 0.929
p̂ = 0.929

Example:

We know from previous chapters that a z score of

3.21 is “unusual” (because it is greater than 2).

It appears that in addition to being greater than

0.5, the sample proportion of 13/14 or 0.929 is

significantly greater than 0.5.

The figure on the next slide shows that the sample

proportion of 0.929 does fall within the range of

values considered to be significant because they

are so far above 0.5 that they are not likely to

occur by chance (assuming that the population

proportion is p = 0.5).

z =
p̂ − p

=
0.929 − 0.5

0.5( ) 0.5( )
14

= 3.21

Example:

Sample proportion of:
or
Test Statistic z = 3.21

p̂ = 0.929

Fail to reject H0 Reject H0

Critical Region

The critical region (or rejection region) is the

set of all values of the test statistic that

cause us to reject the null hypothesis. For

example, see the red-shaded region in the

previous figure.

Significance Level

The significance level (denoted by ) is the

probability that the test statistic will fall in the

critical region when the null hypothesis is

actually true. This is the same  introduced

in Section 7-2. Common choices for  are

0.05, 0.01, and 0.10.

Critical Value

A critical value is any value that separates the

critical region (where we reject the null

hypothesis) from the values of the test

statistic that do not lead to rejection of the null

hypothesis. The critical values depend on the

nature of the null hypothesis, the sampling

distribution that applies, and the significance

level . See the previous figure where the

critical value of z = 1.645 corresponds to a

significance level of  = 0.05.

P-Value

The P-value (or p-value or probability value)

is the probability of getting a value of the test

statistic that is at least as extreme as the one

representing the sample data, assuming that

the null hypothesis is true.

Critical region

in the left tail:

Critical region

in the right tail:

Critical region

in two tails:

P-value = area to the left of

the test statistic

P-value = area to the right of

the test statistic

P-value = twice the area in the

tail beyond the test statistic

P-Value

The null hypothesis is rejected if the P-value

is very small, such as 0.05 or less.

Here is a memory tool useful for interpreting

the P-value:

If the P is low, the null must go.

If the P is high, the null will fly.

Procedure for Finding P-Values

Figure 8-5

H1:>H1:<

H1:≠

Caution

Don’t confuse a P-value with a proportion p.

Know this distinction:

P-value = probability of getting a test

statistic at least as extreme as

the one representing sample

data

p = population proportion

Example

Consider the claim that with the XSORT method

of gender selection, the likelihood of having a

baby girl is different from p = 0.5, and use the

test statistic z = 3.21 found from 13 girls in 14

births.

First determine whether the given conditions

result in a critical region in the right tail, left tail,

or two tails, then use Figure 8-5 to find the P-

value. Interpret the P-value.

H0: p=0.5

H1: p≠0.5

Example

The claim that the likelihood of having a baby

girl is different from p = 0.5 can be expressed as

p ≠ 0.5 so the critical region is in two tails.

Using Figure 8-5 to find the P-value for a two-

tailed test, we see that the P-value is twice the

area to the right of the test statistic z = 3.21.

We refer to Table A-2 (or use technology) to find

that the area to the right of z = 3.21 is 0.0007.

In this case, the P-value is twice the area to the

right of the test statistic, so we have:

P-value = 2  0.0007 = 0.0014

Example

The P-value is 0.0014 (or 0.0013 if greater

precision is used for the calculations). The small

P-value of 0.0014 shows that there is a very

small chance of getting the sample results that

led to a test statistic of z = 3.21. This suggests

that with the XSORT method of gender

selection, the likelihood of having a baby girl is

different from 0.5.

Types of Hypothesis Tests:

Two-tailed, Left-tailed, Right-tailed

The tails in a distribution are the extreme

regions bounded by critical values.

Determinations of P-values and critical values

are affected by whether a critical region is in

two tails, the left tail, or the right tail. It

therefore becomes important to correctly

characterize a hypothesis test as two-tailed,

left-tailed, or right-tailed.

Two-tailed Test

H0: =

H1: 

 is divided equally between
the two tails of the critical

region

Means less than or greater than

Left-tailed Test

H0: =

H1: <

Points Left

 the left tail

Right-tailed Test

H0: =

H1: >

Points Right

 the Right tail

Conclusions

in Hypothesis Testing

We always test the null hypothesis.

The initial conclusion will always be

one of the following:

1. Reject the null hypothesis.

2. Fail to reject the null hypothesis.

P-value method:

Using the significance level :

If P-value   ,

reject H0.

If P-value >  , fail to reject H0.

Decision Criterion

Traditional method:

If the test statistic falls within the

critical region, reject H0.

If the test statistic does not fall

within the critical region, fail to

reject H0.

Decision Criterion

Another option:

Instead of using a significance

level such as 0.05, simply identify

the P-value and leave the decision

to the reader.

Decision Criterion

Confidence Intervals:

A confidence interval estimate of a

population parameter contains the

likely values of that parameter.

If a confidence interval does not

include a claimed value of a

population parameter, reject that

claim.

Wording of Final Conclusion

Figure 8-7

Caution

Never conclude a hypothesis test with a

statement of “reject the null hypothesis”

or “fail to reject the null hypothesis.”

Always make sense of the conclusion

with a statement that uses simple

nontechnical wording that addresses the

original claim.

Accept Versus Fail to Reject

• Some texts use “accept the null
hypothesis.”

• We are not proving the null hypothesis.

• Fail to reject says more correctly

• The available evidence is not strong
enough to warrant rejection of the null
hypothesis (such as not enough
evidence to convict a suspect).

Type I Error

• A Type I error is the mistake of

rejecting the null hypothesis when it

is actually true.

• The symbol  (alpha) is used to

represent the probability of a type I

error.

Type II Error

• A Type II error is the mistake of failing

to reject the null hypothesis when it is

actually false.

• The symbol  (beta) is used to

represent the probability of a type II

error.

Type I and Type II Errors

Example:

a) Identify a type I error.

b) Identify a type II error.

Assume that we are conducting a hypothesis

test of the claim that a method of gender

selection increases the likelihood of a baby

girl, so that the probability of a baby girls is p >

0.5. Here are the null and alternative

hypotheses: H0: p = 0.5, and H1: p > 0.5.

Example:

a) A type I error is the mistake of rejecting a

true null hypothesis, so this is a type I error:

Conclude that there is sufficient evidence to

support p > 0.5, when in reality p = 0.5.

b) A type II error is the mistake of failing to

reject the null hypothesis when it is false, so

this is a type II error: Fail to reject p = 0.5

(and therefore fail to support p > 0.5) when in

reality p > 0.5.

Controlling Type I and

Type II Errors

• For any fixed , an increase in the sample
size n will cause a decrease in 

• For any fixed sample size n, a decrease in
 will cause an increase in . Conversely,
an increase in  will cause a decrease in
.

• To decrease both  and , increase the
sample size.

Comprehensive

Hypothesis Test –

P-Value Method

Comprehensive

Hypothesis Test –

Traditional Method

Comprehensive

Hypothesis Test – cont

A confidence interval estimate of a population

parameter contains the likely values of that

parameter. We should therefore reject a claim

that the population parameter has a value that

is not included in the confidence interval.

In some cases, a conclusion based on a

confidence interval may be different

from a conclusion based on a

hypothesis test. See the comments in

the individual sections.

Caution

Part 2:

Beyond the Basics of
Hypothesis Testing:

The Power of a Test

Definition

The power of a hypothesis test is the

probability (1 –  ) of rejecting a false null

hypothesis. The value of the power is

computed by using a particular significance

level  and a particular value of the

population parameter that is an alternative to

the value assumed true in the null hypothesis.

That is, the power of the hypothesis test is the

probability of supporting an alternative

hypothesis that is true.

Power and the

Design of Experiments

Just as 0.05 is a common choice for a significance

level, a power of at least 0.80 is a common

requirement for determining that a hypothesis test is

effective. (Some statisticians argue that the power

should be higher, such as 0.85 or 0.90.) When

designing an experiment, we might consider how

much of a difference between the claimed value of a

parameter and its true value is an important amount

of difference. When designing an experiment, a goal

of having a power value of at least 0.80 can often be

used to determine the minimum required sample

size.

Recap

In this section we have discussed:

❖ Null and alternative hypotheses.

❖ Test statistics.

❖ Significance levels.

❖ P-values.

❖ Decision criteria.

❖ Type I and II errors.

❖ Power of a hypothesis test.

Section 8-3

Testing a Claim About a

Proportion

Key Concept

This section presents complete procedures

for testing a hypothesis (or claim) made about

a population proportion.

This section uses the components introduced

in the previous section for the P-value

method, the traditional method or the use of

confidence intervals.

Key Concept

Two common methods for testing a claim

about a population proportion are (1) to use a

normal distribution as an approximation to

the binomial distribution, and (2) to use an

exact method based on the binomial

probability distribution.

Part 1 of this section uses the approximate

method with the normal distribution, and Part

2 of this section briefly describes the exact

method.

Part 1:

Basic Methods of Testing Claims
about a Population Proportion p

Notation

p = population proportion (used in the

null hypothesis)

q = 1 – p



n = number of trials

p = (sample proportion)

1) The sample observations are a simple
random sample.

2) The conditions for a binomial distribution
are satisfied.

3) The conditions np  5 and nq  5 are both
satisfied, so the binomial distribution of
sample proportions can be approximated
by a normal distribution with µ = np and
 = npq . Note: p is the assumed
proportion not the sample

proportion.

Requirements for Testing Claims

About a Population Proportion p

p – p

pq
n

z =



Test Statistic for Testing

a Claim About a Proportion

P-values:

Critical Values:

Use the standard normal

distribution (Table A-2) and refer to

Figure 8-5

Use the standard normal

distribution (Table A-2).

Caution

Don’t confuse a P-value with a proportion p.

P-value = probability of getting a test

statistic at least as extreme as

the one representing sample

data

p = population proportion

P-Value Method:

Use the same method as described

in Section 8-2 and in Figure 8-8.

Use the standard normal

distribution (Table A-2).

Traditional Method

Use the same method as described

in Section 8-2 and in Figure 8-9.

Confidence Interval Method

Use the same method as described

in Section 8-2 and in Table 8-2.

CAUTION
When testing claims about a population proportion,

the traditional method and the P-value method are

equivalent and will yield the same result since they

use the same standard deviation based on the claimed

proportion p.

However, the confidence interval uses an estimated

standard deviation based upon the sample proportion p.

Consequently, it is possible that the traditional and P-

value methods may yield a different conclusion than the

confidence interval method.

A good strategy is to use a confidence interval to

estimate a population proportion, but use the P-value

or traditional method for testing a claim about the

proportion.



Example:

The text refers to a study in which 57 out of

104 pregnant women correctly guessed the

sex of their babies.

Use these sample data to test the claim that

the success rate of such guesses is the 50%

success rate expected with random chance

guesses.

Use a 0.05 significance level.

Example:

Requirements are satisfied: simple random

sample; fixed number of trials (104) with two

categories (guess correctly or do not); np =

(104)(0.5) = 52 ≥ 5 and nq = (104)(0.5) = 52 ≥ 5

Step 1: original claim is that the success rate

is no different from 50%: p = 0.50

Step 2: opposite of original claim is p ≠ 0.50

Step 3: p ≠ 0.50 does not contain equality so

it is H1.

H0: p = 0.50 null hypothesis and original claim

H1: p ≠ 0.50 alternative hypothesis

Example:

Step 4: significance level is  = 0.05

Step 5: sample involves proportion so the

relevant statistic is the sample

proportion,

Step 6: calculate z:

p̂

z =
p̂ − p

104
− 0.50

0.50( ) 0.50( )
104

= 0.98

two-tailed test, P-value is twice the

area to the right of test statistic

Example:

Table A-2: z = 0.98 has an area of 0.8365 to its

left, so area to the right is 1 – 0.8365 = 0.1635,

doubles yields 0.3270 (technology provides a

more accurate P-value of 0.3268

Step 7: the P-value of 0.3270 is greater than

the significance level of 0.05, so fail to

reject the null hypothesis

Here is the correct conclusion: There is not

sufficient evidence to warrant rejection of the

claim that women who guess the sex of their

babies have a success rate equal to 50%.

(determining the sample proportion of households with cable TV)

p = = = 0.64
x
n

(96+54)



and 54 do not” is calculated using

p sometimes must be calculated
“96 surveyed households have cable TV




p sometimes is given directly

“10% of the observed sports cars are red”
is expressed as

p = 0.10


Obtaining P


Part 2:

Exact Method for Testing Claims
about a Proportion p

Testing Claims

We can get exact results by using the binomial
probability distribution.

Binomial probabilities are a nuisance to calculate
manually, but technology makes this approach
quite simple.

Also, this exact approach does not require that np
≥ 5 and nq ≥ 5 so we have a method that applies
when that requirement is not satisfied.

To test hypotheses using the exact binomial
distribution, use the binomial probability
distribution with the P-value method, use the value
of p assumed in the null hypothesis, and find P-
values as follows:

Testing Claims

Left-tailed test:

The P-value is the probability of getting

or fewer
successes among n trials.

Right-tailed test:

The P-value is the probability of getting x or more
successes among n trials.

p p̂

Testing Claims

Two-tailed test:

the P-value is twice the probability of
getting x or more successes

the P-value is twice the probability of
getting x or fewer successes

If p̂  p,

If p̂  p,

p p
^

Recap

In this section we have discussed:

❖ Test statistics for claims about a proportion.

❖ P-value method.

❖ Confidence interval method.

❖ Obtaining p.


Section 8-4

Testing a Claim About a

Mean:  Known

Key Concept

This section presents methods for testing a

claim about a population mean, given that the

population standard deviation is a known

value.

This section uses the normal distribution with

the same components of hypothesis tests

that were introduced in Section 8-2.

Notation

n = sample size

= sample mean

= population mean of all sample
means from samples of size n

 = known value of the population
standard deviation

x

Requirements for Testing Claims About

a Population Mean (with  Known)

1) The sample is a simple random
sample.

2) The value of the population standard
deviation  is known.

3) Either or both of these conditions is
satisfied: The population is normally
distributed or

n > 30.

Test Statistic for Testing a Claim
About a Mean (with  Known)

x – µxz =


Example:
People have died in boat accidents because an

obsolete estimate of the mean weight of men was

used.

Using the weights of the simple random sample of

men, we obtain these sample statistics:

n = 40 and = 172.55 lb.

Research from several other sources suggests

that the population of weights of men has a

standard deviation given by  = 26 lb.

Use these results to test the claim that men have a

mean weight greater than 166.3 lb, which was the

weight in the National Transportation and Safety

Board’s

recommendation.

Use a 0.05 significance level, and use the P-value

method outlined in Figure 8-8.

Example:

Requirements are satisfied: simple random

sample,  is known (26 lb), sample size is 40

(n > 30)

Step 1: Express claim as  > 166.3 lb

Step 2: alternative to claim is  ≤ 166.3 lb

Step 3:  > 166.3 lb does not contain equality,

it is the alternative hypothesis:

H0:  = 166.3 lb null hypothesis

H1:  > 166.3 lb alternative hypothesis and

original claim

Example:

Step 4: significance level is  = 0.05

Step 5: claim is about the population mean,

so the relevant statistic is the sample

mean (172.55 lb),  is known (26 lb),

sample size greater than 30

Step 6: calculate z

z =
x − 



=
172.55 − 166.3

= 1.52

right-tailed test, so P-value is the area

to the right of z = 1.52;

Example:

Table A-2: area to the left of z = 1.52

is 0.9357, so the area to the right is

1 – 0.9357 = 0.0643.

The P-value is 0.0643

Step 7: The P-value of 0.0643 is greater than

the significance level of  = 0.05, we

fail

to reject the null hypothesis.

x = 172.55

 = 166.3
or

z = 0 or
z = 1.52

P-value = 0.0643

Example:

The P-value of 0.0643 tells us that if men have a

mean weight given by  = 166.3 lb, there is a

good chance (0.0643) of getting a sample mean

of 172.55 lb.

A sample mean such as 172.55 lb could easily

occur by chance.

There is not sufficient evidence to support a

conclusion that the population mean is greater

than 166.3 lb, as in the National Transportation

and Safety Board’s recommendation.

Example:

The traditional method: Use z = 1.645 instead of

finding the P-value. Since z = 1.52 does not fall

in the critical region, again fail to reject the null

hypothesis.

Confidence Interval method: Use a one-tailed

test with  = 0.05, so construct a 90%

confidence interval:

165.8 <  < 179.3

The confidence interval contains 166.3 lb, we

cannot support a claim that  is greater than

166.3. Again, fail to reject the null hypothesis.

Underlying Rationale of

Hypothesis Testing

If, under a given assumption, there is an

extremely small probability of getting sample

results at least as extreme as the results that

were obtained, we conclude that the

assumption is probably not correct.

When testing a claim, we make an

assumption (null hypothesis) of equality. We

then compare the assumption and the

sample results and we form one of the

following conclusions:

• If the sample results (or more extreme results)

can easily occur when the assumption (null

hypothesis) is true, we attribute the relatively

small discrepancy between the assumption and

the sample results to chance.

• If the sample results cannot easily occur when

that assumption (null hypothesis) is true, we

explain the relatively large discrepancy between

the assumption and the sample results by

concluding that the assumption is not true, so

we reject the assumption.

Underlying Rationale of

Hypotheses Testing – cont

Recap

In this section we have discussed:

❖ Requirements for testing claims about

population means, σ known.

❖ P-value method.

❖ Traditional method.

❖ Confidence interval method.

❖ Rationale for hypothesis testing.

Section 8-5

Testing a Claim About a

Mean:  Not Known

Key Concept

This section presents methods for testing a

claim about a population mean when we do

not know the value of σ.

The methods of this section use the Student t

distribution introduced earlier.

Notation

n = sample size

= sample mean

= population mean of all sample
means from samples of size n

x

Requirements for Testing Claims

About a Population

Mean (with  Not Known)

1) The sample is a simple random sample.

2) The value of the population standard
deviation  is not known.

3) Either or both of these conditions is
satisfied: The population is normally
distributed or n > 30.

Test Statistic for Testing a
Claim About a Mean
(with  Not Known)

P-values and Critical Values
❖Found in Table A-3

❖Degrees of freedom (df) = n – 1

x – µxt = s
n

Important Properties of the

Student t Distribution

1. The Student t distribution is different for different

sample sizes (see Figure 7-5 in Section 7-4).

2. The Student t distribution has the same general bell
shape as the normal distribution; its wider shape
reflects the greater variability that is expected when s is
used to estimate  .

3. The Student t distribution has a mean of t = 0 (just as
the standard normal distribution has a mean of z = 0).

4. The standard deviation of the Student t distribution
varies with the sample size and is greater than 1 (unlike
the standard normal distribution, which has  = 1).

5. As the sample size n gets larger, the Student t
distribution gets closer to the standard normal
distribution.

Choosing between the Normal and
Student t Distributions when Testing a

Claim about a Population Mean µ

Use the Student t distribution when  is

not known and either or both of these

conditions is satisfied:

The population is normally distributed or

n > 30.

Example:
People have died in boat accidents because an

obsolete estimate of the mean weight of men was

used.

Using the weights of the simple random sample

of men from Data Set 1 in Appendix B, we obtain

these sample statistics: n = 40 and = 172.55 lb,

and s= 26.33 lb. Do not assume that the value of 

is known.

Use these results to test the claim that men have a

mean weight greater than 166.3 lb, which was the

weight in the National Transportation and Safety

Board’s recommendation M-04-04.

Use a 0.05 significance level, and the traditional

method outlined in Figure 8-9.

Example:

Requirements are satisfied: simple random

sample, population standard deviation is not

known, sample size is 40 (n > 30)

Step 1: Express claim as  > 166.3 lb

Step 2: alternative to claim is  ≤ 166.3 lb

Step 3:  > 166.3 lb does not contain equality,

it is the alternative hypothesis:

H0:  = 166.3 lb null hypothesis

H1:  > 166.3 lb alternative hypothesis and

original claim

Example:

Step 4: significance level is  = 0.05

Step 5: claim is about the population mean,

so the relevant statistic is the sample

mean, 172.55 lb

Step 6: calculate t

t =
x − 

=
172.55 − 166.3

26.33

= 1.501

df = n – 1 = 39, area of 0.05, one-tail

yields t = 1.685;

Example:

Step 7: t = 1.501 does not fall in the critical

region bounded by t = 1.685, we fail

to reject the null hypothesis.

 = 166.3
or

z = 0
x = 172.55

or
t = 1.52

Critical value
t = 1.685

Example:

Because we fail to reject the null hypothesis, we

conclude that there is not sufficient evidence to

support a conclusion that the population mean

is greater than 166.3 lb, as in the National

Transportation and Safety Board’s

recommendation.

The critical value in the preceding example

was t = 1.782, but if the normal distribution

were being used, the critical value would have

been z = 1.645.

The Student t critical value is larger (farther to

the right), showing that with the Student t

distribution, the sample evidence must be

more extreme before we can consider it to be

significant.

Normal Distribution Versus

Student t Distribution

P-Value Method

❖ Use software or a TI-83/84 Plus
calculator.

❖ If technology is not available, use Table
A-3 to identify a range of P-values.

a) In a left-tailed hypothesis test, the sample size

is n = 12, and the test statistic is t = –2.007.

b) In a right-tailed hypothesis test, the sample size

is n = 12, and the test statistic is t = 1.222.

c) In a two-tailed hypothesis test, the sample size

is n = 12, and the test statistic is t = –3.456.

Example: Assuming that neither software nor
a TI-83 Plus calculator is available, use Table
A-3 to find a range of values for the P-value
corresponding to the given results.

a) The test is a left-tailed test with test

statistic t = –2.007, so the P-value is the

area to the left of –2.007. Because of the

symmetry of the t distribution, that is the

same as the area to the right of +2.007. Any

test statistic between 2.201 and 1.796 has a

right-tailed P-value that is between 0.025

and 0.05. We conclude that

0.025 < P-value < 0.05.

Example: Assuming that neither software nor
a TI-83 Plus calculator is available, use Table
A-3 to find a range of values for the P-value
corresponding to the given results.

b) The test is a right-tailed test with test

statistic t = 1.222, so the P-value is the

area to the right of 1.222. Any test

statistic less than 1.363 has a right-tailed

P-value that is greater than 0.10. We

conclude that P-value > 0.10.

c) The test is a two-tailed test with test statistic

t = –3.456. The P-value is twice the area to

the left of –3.456 (= right of +3.456). Any test

statistic greater than 3.106 has a two-

tailed P-value that is less than 0.01. We

conclude that

P-value < 0.01.

Example: Assuming that neither software nor
a TI-83 Plus calculator is available, use Table
A-3 to find a range of values for the P-value
corresponding to the given results.

Recap

In this section we have discussed:

❖ Assumptions for testing claims about

population means, σ unknown.

❖ Student t distribution.

❖ P-value method.

Lecture Slides

Elementary Statistics
Eleventh Edition

and the Triola Statistics Series

by Mario F. Triola

Chapter 6

Normal Probability

Distribution

6-1

Review

and Preview

6-2 The

Standard

Normal Distribution

6-3 Applications of Normal

Distributions

6-4 Sampling Distributions

and Estimators

6-5 The Central Limit

Theorem

Section 6-1

Review and

Preview

❖ Chapter 2: Distribution of data

❖ Chapter 3: Measures of data sets,

including measures of center and

variation

❖ Chapter 4: Principles of probability

❖ Chapter 5: Discrete probability

distributions

Review

Chapter focus is on:

❖ Continuous random variables

❖ Normal distributions

Preview

Figure 6-1

Formula 6-1

( )=
e
−
1

x−










 2

Distribution determined

by fixed values of mean

and standard deviation

Section 6-2

The Standard Normal

Distribution

Key Concept

This section presents the standard

normal

distribution which has three properties:

1. It’s graph is bell-shaped.

2. It’s mean is equal to 0 ( = 0).

3. It’s standard deviation is equal to 1 ( = 1).

Develop the skill to find areas (or probabilities

or relative frequencies) corresponding to

various regions under the graph of the

standard normal

distribution.

Find z-scores that

correspond to area under the graph.

Uniform Distribution

A continuous random variable has a

uniform distribution if its values are

spread evenly over the range of

probabilities. The graph of a uniform

distribution results in a rectangular shape.

A density curve is the graph of a
continuous probability

distribution.

It
must satisfy the following properties:

Density Curve

1. The total area under the curve must
equal 1.

2. Every point on the curve must have a
vertical height that is 0 or greater.
(That is, the curve cannot fall below
the x-axis.)

Because the total area under the

density curve is equal to 1,

there is a correspondence

between area and probability.

Area and Probability

Using Area to Find Probability

Given the uniform distribution illustrated, find
the probability that a randomly selected
voltage level is greater than 124.5 volts.

Shaded area

represents

voltage levels

greater than

124.5 volts.

Correspondence

between area

and probability:

0.25.

Standard Normal Distribution

The standard normal distribution is a

normal probability distribution with  = 0

and  = 1. The total area under its density

curve is equal to 1.

Finding Probabilities When

Given z-scores

❖

Table A-2

(in Appendix A)

❖ Formulas and Tables insert card

❖ Find areas for many different
regions

Finding Probabilities –

Other Methods

❖ STATDISK

❖ Minitab

❖ Excel

❖ TI-83/84 Plus

Methods for Finding Normal

Distribution Areas

Methods for Finding Normal

Distribution Areas

Table A-2

1. It is designed only for the standard normal
distribution, which has a mean of 0 and a
standard deviation of 1.

2. It is on two pages, with one page for negative
z-scores and the other page for positive
z-scores.

3. Each value in the body of the table is a
cumulative area from the left up to a vertical
boundary above a specific z-score.

Using Table A-2

4. When working with a graph, avoid confusion

between z-scores and areas.

z Score

Distance along horizontal scale of the

standard normal distribution; refer to the

leftmost column and top row of Table A-2.

Area

Region under the curve; refer to the values in

the body of Table A-2.

5. The part of the z-score denoting hundredths

is found across the top.

Using Table A-2

The Precision Scientific Instrument Company
manufactures thermometers that are supposed
to give readings of 0ºC at the freezing point of
water. Tests on a large sample of these
instruments reveal that at the freezing point of
water, some thermometers give readings below
0º (denoted by negative numbers) and some give
readings above 0º (denoted by positive
numbers). Assume that the mean reading is 0ºC
and the standard deviation of the readings is
1.00ºC. Also assume that the readings are
normally distributed. If one thermometer is
randomly selected, find the probability that, at
the freezing point of water, the reading is less
than 1.27º.

Example – Thermometers

P(z < 1.27) =

Example – (Continued)

Look at Table A-2

P (z < 1.27) = 0.8980

Example – cont

The probability of randomly selecting a
thermometer with a reading less than 1.27º
is 0.8980.

P (z < 1.27) = 0.8980

Example – cont

Or 89.80% will have readings below 1.27º.

P (z < 1.27) = 0.8980 Example - cont

If thermometers have an average (mean) reading of 0
degrees and a standard deviation of 1 degree for
freezing water, and if one thermometer is randomly
selected, find the probability that it reads (at the
freezing point of water) above –1.23 degrees.

Probability of randomly selecting a thermometer
with a reading above –1.23º is 0.8907.

P (z > –1.23) = 0.8907

Example – Thermometers Again

P (z > –1.23) = 0.8907

89.07% of the thermometers have readings
above –1.23 degrees.

Example – cont

A thermometer is randomly selected. Find the probability
that it reads (at the freezing point of water) between –2.00
and 1.50 degrees.

P (z < –2.00) = 0.0228 P (z < 1.50) = 0.9332 P (–2.00 < z < 1.50) = 0.9332 – 0.0228 = 0.9104

The probability that the chosen thermometer has a

reading between – 2.00 and 1.50 degrees is 0.9104.

Example – Thermometers III

If many thermometers are selected and tested at the
freezing point of water, then 91.04% of them will read
between –2.00 and 1.50 degrees.

P (z < –2.00) = 0.0228 P (z < 1.50) = 0.9332 P (–2.00 < z < 1.50) = 0.9332 – 0.0228 = 0.9104 A thermometer is randomly selected. Find the probability that it reads (at the freezing point of water) between –2.00 and 1.50 degrees. Example - cont

P(a < z < b) denotes the probability that the z score is between a and b.

P(z > a)

denotes the probability that the z score is greater than a.

P(z < a) denotes the probability that the z score is less than a.

Notation

Finding a z Score When Given a

Probability Using Table A-2

1. Draw a bell-shaped curve and identify the region
under the curve that corresponds to the given
probability. If that region is not a cumulative
region from the left, work instead with a known
region that is a cumulative region from the left.

2. Using the cumulative area from the left, locate the
closest probability in the body of Table A-2 and
identify the corresponding z score.

Finding z Scores
When Given Probabilities

5% or 0.05

(z score will be positive)

Finding the 95th Percentile

Finding z Scores
When Given Probabilities – cont

Finding the 95th Percentile

1.645

5% or 0.05

(z score will be positive)

Finding the Bottom 2.5% and Upper 2.5%

(One z score will be negative and the other positive)

Finding z Scores
When Given Probabilities – cont

Finding the Bottom 2.5% and Upper 2.5%

(One z score will be negative and the other positive)

Finding z Scores
When Given Probabilities – cont

Finding the Bottom 2.5% and Upper 2.5%

(One z score will be negative and the other positive)

Finding z Scores
When Given Probabilities – cont

Recap

In this section we have discussed:

❖ Density curves.

❖ Relationship between area and probability.

❖ Standard normal

distribution.

❖ Using Table A-2.

Section 6-3

Applications of Normal

Distributions

Key Concept

This section presents methods for working

with normal distributions that are not standard.

That is, the mean is not 0 or the standard

deviation is not 1, or both.

The key concept is that we can use a simple

conversion that allows us to standardize any

normal distribution so that the same methods

of the previous section can be used.

Conversion Formula

x – µ
z =

Round z scores to 2 decimal places

Converting to a Standard

Normal Distribution

x – 


z =

In the Chapter Problem, we noted that the safe

load for a water taxi was found to be 3500

pounds. We also noted that the mean weight of a

passenger was assumed to be 140 pounds.

Assume the worst case that all passengers are

men. Assume also that the weights of the men

are normally distributed with a mean of 172

pounds and standard deviation of 29 pounds. If

one man is randomly selected, what is the

probability he weighs less than 174 pounds?

Example – Weights of

Water Taxi Passengers

Example – cont

z =
174 – 172

29
= 0.07

 = 29

 = 172

Example – cont

P ( x < 174 lb.) = P(z < 0.07)

= 0.5279

 = 29

 = 172

1. Don’t confuse z scores and areas. z scores are

distances along the horizontal scale, but areas

are regions under the normal curve. Table A-2

lists z scores in the left column and across the top

row, but areas are found in the body of the table.

2. Choose the correct (right/left) side of the graph.

3. A z score must be negative whenever it is located

in the left half of the

normal distribution.

4. Areas (or probabilities) are positive or zero values,

but they are never negative.

Helpful Hints

Procedure for Finding Values

Using Table A-2 and Formula 6-2
1. Sketch a normal distribution curve, enter the given probability or

percentage in the appropriate region of the graph, and identify
the x value(s) being sought.

2. Use Table A-2 to find the z score corresponding to the cumulative
left area bounded by x. Refer to the body of Table A-2 to find the
closest area, then identify the corresponding z score.

3. Using Formula 6-2, enter the values for µ, , and the z score
found in step 2, then solve for x.

x = µ + (z • ) (Another form of Formula 6-2)

(If z is located to the left of the mean, be sure that it is a negative
number.)

4. Refer to the sketch of the curve to verify that the solution makes
sense in the context of the graph and the context of the problem.

Example – Lightest and Heaviest

Use the data from the previous example to determine

what weight separates the lightest 99.5% from the

heaviest 0.5%?

x =  + (z ● )

x = 172 + (2.575 • 29)

x = 246.675 (247 rounded)

Example –

Lightest and Heaviest – cont

The weight of 247 pounds separates the

lightest 99.5% from the heaviest 0.5%

Example –

Lightest and Heaviest – cont

Recap

In this section we have discussed:

❖ Non-standard normal distribution.

❖ Converting to a standard normal distribution.

❖ Procedures for finding values using Table A-2

and Formula 6-2.

Section 6-4

Sampling Distributions

and Estimators

Key Concept

The main objective of this section is to

understand the concept of a sampling

distribution of a statistic, which is the

distribution of all values of that statistic

when all possible samples of the same size

are taken from the same population.

We will also see that some statistics are

better than others for estimating population

parameters.

Definition

The sampling distribution of a statistic (such

as the sample mean or sample proportion) is

the distribution of all values of the statistic

when all possible samples of the same size n

are

taken from the same population. (The

sampling distribution of a statistic is typically

represented as a probability distribution in the

format of a table, probability histogram, or

formula.)

Definition

The

sampling distribution of the mean is

the distribution of sample means, with all

samples having the same sample size n

taken from the same population. (The

sampling distribution of the mean is

typically represented as a probability

distribution in the format of a table,

probability histogram, or formula.)

Properties

❖ Sample means target the value of the

population mean. (That is, the mean of the

sample means is the population mean. The

expected value of the sample mean is equal

to the population mean.)

❖ The distribution of the sample means tends

to be a normal distribution.

Definition

The sampling distribution of the variance is the

distribution of sample variances, with all

samples having the same sample size n taken

from the same population. (The sampling

distribution of the variance is typically

represented as a probability distribution in the

format of a table, probability histogram, or

formula.)

Properties

❖ Sample variances target the value of the

population variance. (That is, the mean of

the sample variances is the population

variance. The expected value of the sample

variance is equal to the population variance.)

❖ The distribution of the sample variances

tends to be a distribution skewed to the

right.

Definition

The sampling distribution of the proportion

is the distribution of sample proportions,

with all samples having the same sample

size n taken from the same population.

Definition

We need to distinguish between a

population proportion p and some sample

proportion:

p = population proportion

= sample proportion p̂

Properties

❖ Sample proportions target the value of the

population proportion. (That is, the mean of

the sample proportions is the population

proportion. The expected value of the

sample proportion is equal to the population

proportion.)

❖ The distribution of the sample proportion

tends to be a normal distribution.

Unbiased Estimators

Sample means, variances and

proportions are unbiased estimators.

That is they target the population

parameter.

These statistics are better in estimating

the population parameter.

Biased Estimators

Sample medians, ranges and standard

deviations are biased estimators.

That is they do NOT target the

population parameter.

Note: the bias with the standard

deviation is relatively small in large

samples so s is often used to estimate.

Example – Sampling Distributions

Consider repeating this process: Roll a die 5

times, find the mean , variance s2, and the

proportion of odd numbers of the

results.

What do we know about the behavior of all

sample means that are generated as this

process continues indefinitely?

Example – Sampling Distributions

All outcomes are equally likely so the

population mean is 3.5; the mean of the

10,000 trials is 3.49. If continued indefinitely,

the sample mean will be 3.5. Also, notice the

distribution is “normal.”

Specific results from 10,000 trials

Example – Sampling Distributions

All outcomes are equally likely so the

population variance is 2.9; the mean of the

10,000 trials is 2.88. If continued indefinitely,

the sample variance will be 2.9. Also, notice

the distribution is “skewed to the right.”

Specific results from 10,000 trials

Example – Sampling Distributions

All outcomes are equally likely so the

population proportion of odd numbers is 0.50;

the proportion of the 10,000 trials is 0.50. If

continued indefinitely, the mean of sample

proportions will be 0.50. Also, notice the

distribution is “approximately normal.”

Specific results from 10,000 trials

Why Sample with Replacement?

Sampling without replacement would have the very

practical advantage of avoiding wasteful duplication

whenever the same item is selected more than once.

However, we are interested in sampling with

replacement for these two reasons:

1. When selecting a relatively small sample form a

large population, it makes no significant

difference whether we sample with replacement

or without replacement.

2. Sampling with replacement results in

independent events that are unaffected by

previous outcomes, and independent events are

easier to analyze and result in simpler

calculations and formulas.

Caution

Many methods of statistics require a simple

random sample. Some samples, such as

voluntary response samples or convenience

samples, could easily result in very wrong

results.

Recap

In this section we have discussed:

❖ Sampling distribution of a statistic.

❖ Sampling distribution of the mean.

❖ Sampling distribution of the variance.

❖ Sampling distribution of the proportion.

❖ Estimators.

Section 6-5

The Central Limit

Theorem

Key Concept

The Central Limit Theorem tells us that for a

population with any distribution, the

distribution of the sample means approaches

a normal distribution as the sample size

increases.

The procedure in this section form the

foundation for estimating population

parameters and hypothesis testing.

Central Limit Theorem

1. The random variable x has a distribution (which may
or may not be normal) with mean µ and standard
deviation .

2. Simple random samples all of size n are selected
from the population. (The samples are selected so
that all possible samples of the same size n have the
same chance of being selected.)

Given:

1. The distribution of sample x will, as the

sample size increases, approach a normal

distribution.

2. The mean of the sample means is the

population mean µ.

3. The standard deviation of all sample means

Conclusions:

Central Limit Theorem – cont.

 n.

Practical Rules Commonly Used

1. For samples of size n larger than 30, the

distribution of the sample means can be

approximated reasonably well by a normal

distribution. The approximation gets closer

to a normal distribution as the sample size n

becomes larger.

2. If the original population is normally

distributed, then for any sample size n, the

sample means will be normally distributed

(not just the values of n larger than 30).

Notation

the mean of the sample means

the standard deviation of sample mean

(often called the standard error of the mean)

µx = µ

nx = 

Example – Normal Distribution

As we proceed

from n = 1 to

n = 50, we see

that the

distribution of

sample means

is approaching

the shape of a

normal

distribution.

Example – Uniform Distribution

As we proceed

from n = 1 to

n = 50, we see

that the

distribution of

sample means

is approaching

the shape of a

normal

distribution.

Example – U-Shaped Distribution

As we proceed

from n = 1 to

n = 50, we see

that the

distribution of

sample means

is approaching

the shape of a

normal

distribution.

As the sample size increases, the

sampling distribution of sample

means approaches a normal

distribution.

Important Point

Use the Chapter Problem. Assume the
population of weights of men is normally
distributed with a mean of 172 lb and a
standard deviation of 29 lb.

Example – Water Taxi Safety

a) Find the probability that if an individual man
is randomly selected, his weight is greater
than 175 lb.

b) b) Find the probability that 20 randomly
selected men will have a mean weight that is
greater than 175 lb (so that their total weight
exceeds the safe capacity of 3500 pounds).

z = 175 – 172 = 0.10
29

a) Find the probability that if an individual man
is randomly selected, his weight is greater
than 175 lb.

Example – cont

b) Find the probability that 20 randomly
selected men will have a mean weight that is
greater than 175 lb (so that their total weight
exceeds the safe capacity of 3500 pounds).

Example – cont

z = 175 – 172 = 0.46
29

b) Find the probability that 20 randomly selected men

will have a mean weight that is greater than 175 lb (so

that their total weight exceeds the safe capacity of

3500 pounds).

It is much easier for an individual to deviate from the
mean than it is for a group of 20 to deviate from the mean.

a) Find the probability that if an individual man is

randomly selected, his weight is greater than 175 lb.

Example – cont

P(x > 175) = 0.4602

P(x > 175) = 0.3228

Interpretation of Results

Given that the safe capacity of the water taxi

is 3500 pounds, there is a fairly good chance

(with probability 0.3228) that it will be

overloaded with 20 randomly selected men.

Correction for a Finite Population

N – n
x

=

n N – 1

finite population
correction factor

When sampling without replacement and the sample

size n is greater than 5% of the finite population of

size N (that is, n > 0.05N ), adjust the standard

deviation of sample means by multiplying it by the

finite population correction factor:

Recap

In this section we have discussed:

❖ Central limit theorem.

❖ Practical rules.

❖ Effects of sample sizes.

❖ Correction for a finite population.

Section 6-7

Assessing Normality

Key Concept

This section presents criteria for determining

whether the requirement of a normal

distribution is satisfied.

The criteria involve visual inspection of a

histogram to see if it is roughly bell shaped,

identifying any outliers, and constructing a

graph called a normal quantile plot.

Definition

A normal quantile plot (or normal

probability plot) is a graph of points (x,y),

where each x value is from the original set

of sample data, and each y value is the

corresponding z score that is a quantile

value expected from the standard normal

distribution.

Procedure for Determining Whether

It Is Reasonable to Assume that

Sample Data are From a Normally

Distributed Population
1. Histogram: Construct a histogram. Reject

normality if the histogram departs dramatically
from a bell shape.

2. Outliers: Identify outliers. Reject normality if
there is more than one outlier present.

3. Normal Quantile Plot: If the histogram is
basically symmetric and there is at most one
outlier, use technology to generate a normal
quantile plot.

Procedure for Determining Whether

It Is Reasonable to Assume that

Sample Data are From a Normally

Distributed Population
3. Continued

Use the following criteria to determine whether
or not the distribution is normal.

Normal Distribution: The population distribution
is normal if the pattern of the points is
reasonably close to a straight line and the
points do not show some systematic pattern
that is not a

straight-line pattern.

Procedure for Determining Whether

It Is Reasonable to Assume that

Sample Data are From a Normally

Distributed Population
3. Continued

Not a Normal Distribution: The population distribution is

not normal if either or both of these two conditions

applies:

❖ The points do not lie reasonably close to a straight

line.

❖ The points show some systematic pattern that is not a

straight-line pattern.

Example

Normal: Histogram of IQ scores is close to being bell-

shaped, suggests that the IQ scores are from a normal

distribution. The normal quantile plot shows points that

are reasonably close to a straight-line pattern. It is safe to

assume that these IQ scores are from a normally

distributed population.

Example

Uniform: Histogram of data having a uniform distribution.

The corresponding normal quantile plot suggests that the

points are not normally distributed because the points

show a systematic pattern that is not a straight-line

pattern. These sample values are not from a population

having a normal distribution.

Example

Skewed: Histogram of the amounts of rainfall in Boston for

every Monday during one year. The shape of the histogram

is skewed, not bell-shaped. The corresponding normal

quantile plot shows points that are not at all close to a

straight-line pattern. These rainfall amounts are not from a

population having a normal distribution.

Manual Construction of a

Normal Quantile Plot

Step 1. First sort the data by arranging the values in

order from lowest to highest.

Step 2. With a sample of size n, each value represents a

proportion of 1/n of the sample. Using the known

sample size n, identify the areas of 1/2n, 3/2n,

and so on. These are the cumulative areas to the

left of the corresponding sample values.

Step 3. Use the standard normal distribution (Table A-2

or software or a calculator) to find the z scores

corresponding to the cumulative left areas found

in Step 2. (These are the z scores that are

expected from a normally distributed sample.)

Manual Construction of a

Normal Quantile Plot

Step 4. Match the original sorted data values with their

corresponding z scores found in Step 3, then

plot the points (x, y), where each x is an original

sample value and y is the corresponding z score.

Step 5. Examine the normal quantile plot and determine

whether or not the distribution is normal.

Ryan-Joiner Test

The Ryan-Joiner test is one of several formal

tests of normality, each having their own

advantages and disadvantages. STATDISK has a

feature of Normality Assessment that displays a

histogram, normal quantile plot, the number of

potential outliers, and results from the Ryan-

Joiner test. Information about the Ryan-Joiner

test is readily available on the Internet.

Data Transformations

Many data sets have a distribution that is not

normal, but we can transform the data so that

the modified values have a normal distribution.

One common transformation is to replace each

value of x with log (x + 1). If the distribution of

the log (x + 1) values is a normal distribution,

the distribution of the x values is referred to as

a lognormal distribution.

Other Data Transformations

In addition to replacing each x value with the

log (x + 1), there are other transformations,

such as replacing each x value with , or 1/x,

or x2. In addition to getting a required normal

distribution when the original data values are

not normally distributed, such transformations

can be used to correct other deficiencies, such

as a requirement (found in later chapters) that

different data sets have the same variance.

Recap

In this section we have discussed:

❖ Normal quantile plot.

❖ Procedure to determine if data have a

normal distribution.

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