CHAPTER 9
HYPOTHESIS TESTS
ABOUT THE MEAN
Prem Mann, Introductory Statistics, 9/E
Copyright © 2015 John Wiley & Sons. All rights reserved.
Why do we Conduct A Hypothesis Test?
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In a test of hypothesis, we test a certain given theory/
belief/ claim about a population parameter.
By using sample statistic we test whether a given
claim about a population parameter is true.
A soft-drink company may claim that, on average, their
cans contain 12 ounces of soda. A government agency
wants to test whether such cans contain, on average, 12
ounces of soda.
Suppose we take a sample of 100 cans of the soft drink
under investigation. We then find out that the mean
amount of soda in these 100 cans is 11.89 ounces.
Based on this result, can we state that, on
average, all such cans contain less than 12 ounces
of soda and that the company is lying to the
public?
Why do we Conduct A Hypothesis Test?
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Not until we perform a test of hypothesis can we
make such an accusation.
The reason is that the mean of 11.89 ounces, is
obtained from a sample. The difference between 12
ounces and 11.89 ounces may have occurred only
because of the sampling error.
Another sample of 100 cans may give us a mean of
12.04 ounces. Therefore, we perform a test of
hypothesis to find out how large the difference
between 12 ounces and 11.89 ounces is and to
investigate whether this difference has occurred as a
result of chance alone.
9.1 Hypothesis Tests: An Introduction
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Two Hypotheses
Rejection and Non-rejection Regions
Two Types of Errors
Tails of a Test
Two Hypotheses
A null hypothesis is a claim (or statement) about a
population parameter that is assumed to be true until it
is declared false.
Null: The person is not guilty
An alternative hypothesis will be declared true if the
null hypothesis is proved to be false.
Alternative: The person is guilty
Rejection and Non-rejection Regions
Figure: Non-rejection and Rejection Regions for the
Court Case
Do not/Fail to
Reject the Null
Hypothesis
Reject the Null
Hypothesis/The
Alternate
Hypothesis is
True
Two Types of Errors
Table: Four Possible Outcomes for a Court Case
A court’s verdict might not always be correct. If a person is
declared guilty at the end of a trial, there are two possibilities.
Type I error & Type II error
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A Type I error occurs when a true null hypothesis is
rejected.
The value of α represents the probability of committing this type
of error; that is,
α = P(H0 is rejected | H0 is true)
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α is called the significance level of the test.
In one approach to a test of hypothesis, we assign a value to α
before making the test. Although any value can be assigned to
α to the commonly used values are .01, .025, .05, and .10.
Type I error & Type II error
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A Type II error occurs when a false null hypotheses is not
rejected.
The value of β represents the probability of committing a Type II
error; that is,
β = P (H0 is not rejected | H0 is false)
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(1 – β) is called the power of the test. It represents the
probability of not making a Type II error.
Table: Four Possible Outcomes for a Test of
Hypothesis
Tails of a Test
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The hypothesis-testing procedure is like the trial of a
person in court but with two major differences.
First is that in a test of hypothesis, the partition of the
rejection and nonrejection regions is not arbitrary.
Instead, it depends on the value assigned to α or
significance level of the test.
Second difference relates to the rejection region. In
the court case, the rejection region is on the right side
of the critical point. However, in statistics, the rejection
region for a hypothesis-testing problem can be on both
sides, with the non-rejection region in the middle, or it
can be on the left side or right side of the nonrejection
region.
Tails of a Test
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A two-tailed test has rejection regions in both tails.
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A left-tailed test has the rejection region in the left tail.
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A right-tailed test has the rejection region in the right tail
of the distribution curve.
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What tailed test will be conducted is determined by
the sign in the alternative hypothesis.
A Two-Tailed Test
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According to the U.S. Bureau of Statistics, workers in the
United States earn an average of $47,230 a year. Suppose
an economist wants to check whether this mean has changed
or not. The population of US workers is approximately
normally distributed.
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The key word here is changed.
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The test is to check if the income has either increased or
decreased. It is not the same. Which means it is not equal to
the previous income.
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This is an example of a two tailed test.
A Two-Tailed Test
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Let μ be the mean annual earning of employed
Americans. The two possible decisions are

H0 : μ = $47,230 (The mean annual earning of
employed Americans- this is the claim, which will
determine the null hypothesis)

H1 : μ ≠ $47,230 (The mean annual earning of
employed Americans has changed-this is what we
want to test, which will determine the alternate
hypothesis)
A Two-Tailed Test
 Whether a test is two–tailed or one–tailed is
determined by the sign in the alternative
hypothesis.
 If the alternative hypothesis has a not equal
to (≠) sign, it is a two–tailed test.
 A two-tailed test has 2 rejection regions, one
in each tail of the normal distribution curve.
Figure 9.2 A Two-Tailed Test
A Two-Tailed Test
 The curve shows the sampling distribution of
has a normal distribution (bell-shaped-curve)
 The distribution has a mean
 The significance level of
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𝒙, Assuming it
𝝁 of $47,230
𝜶 denotes the total area in the
rejection region.
Since there are 2 rejection regions and the normal
distribution curve is symmetric- there is approximately 𝜶/2
areas on each side of the tail.
There are 2 critical values that denote where the rejection
region begins. (we find these values from the z-table)
The middle area is called the non-rejection region.
We will reject the Null if the value from the sample mean
falls in the rejection region on the 2 ends/tails.
A Two-Tailed Test
 By rejecting the null we are essentially saying that the
difference between the value of 𝝁 stated in the 𝐻0 and
the value of 𝒙 obtained from the sample is too large to
have occurred just out of the sampling error. The
difference in the population and sample is REAL.
 Consequently, by not rejecting the 𝐻0 (if the value falls
in the non-rejection region) we are saying that the
difference between the value of 𝝁 stated in the 𝐻0 and
the value of 𝒙 obtained from the sample is small and
can have occurred from the sampling error.
A Left-Tailed Test

A soda company claims that their cans, on average,
contain 12 ounces of soda. However, if these cans
contain less than the claimed amount of soda, then the
company can be accused of cheating. Suppose a
consumer agency wants to test whether the mean
amount of soda per can is less than 12 ounces.
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Note that the key phrase this time is less than, which
indicates that this is a left-tailed test.
A Left-Tailed Test
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Let μ be the mean amount of soda in all cans. The two
possible decisions are
 H0 : μ = 12 ounces (The mean is equal to 12 ounces,
which the company claims)
 H1 : μ < 12 ounces (The mean is less than 12
ounces, which we want to test)

When the alternative hypothesis has a less than (<)
sign, the test is always left–tailed.
Figure 9.3 A Left-Tailed Test
Prem Mann, Introductory Statistics, 9/E
Copyright © 2015 John Wiley & Sons. All rights reserved.
A Left-Tailed Test
 In a left-tailed test, the rejection region is in the left tail
of the distribution curve, as shown in Figure and the
area of this rejection region is equal to 𝜶 (the
significance level).
 We can observe from this figure that there is only one
critical value in a left-tailed test.
A Right-Tailed Test
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The average price of homes in West Orange, New
Jersey, was $471,257. Suppose a real estate
researcher wants to check whether the current mean
price of homes in this town is higher than $471,257.
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The key phrase in this case is higher than, which
indicates a right-tailed test.
Prem Mann, Introductory Statistics, 9/E
Copyright © 2015 John Wiley & Sons. All rights reserved.
A Right-Tailed Test
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Let μ be the current mean price of homes in this town. The
two possible decisions are
 H0 : μ = $471,257 (The current mean price of homes in
this town is $ 471,257)
 H1 : μ > $471,257 (The current mean price of homes in
this town is higher than $ 471,257)

When the alternative hypothesis has a greater than (>)
sign, the test is always right–tailed.
Prem Mann, Introductory Statistics, 9/E
Copyright © 2015 John Wiley & Sons. All rights reserved.
Figure 9.4 A Right-Tailed Test
Prem Mann, Introductory Statistics, 9/E
Copyright © 2015 John Wiley & Sons. All rights reserved.
A Right-Tailed Test
 In a right-tailed test, the rejection region is in the right
tail of the distribution curve, as shown in Figure and the
area of this rejection region is equal to 𝛼 (the
significance level).
 We can observe from this figure that there is only one
critical value in a right-tailed test.
Prem Mann, Introductory Statistics, 9/E
Copyright © 2015 John Wiley & Sons. All rights reserved.
Table 9.3 Signs in H0 and H1 and Tails of a
Test
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Copyright © 2015 John Wiley & Sons. All rights reserved.
9.2 Hypothesis Tests About µ : σ Known
This section explains how to perform a test of hypothesis for the
population mean when the population standard deviation is
known/given. Three Possible Cases
Case I. If the following three conditions are fulfilled:
1. The population standard deviation σ is known
2. The sample size is small (i.e., n < 30)
3. The population from which the sample is selected is
normally distributed.
We use the normal distribution to perform a test of hypothesis.
Hypothesis Tests About µ : σ Known
Case II. If the following two conditions are fulfilled:
1. The population standard deviation σ is known
2. The sample size is large (i.e., n ≥ 30)
We use the normal distribution to perform a test of hypothesis
through the help of the C.L Theorem.
Case III. If the following three conditions are fulfilled:
1. The population standard deviation σ is known
2. The sample size is small (i.e., n < 30)
3. The population from which the sample is selected is
not normally distributed (or its distribution is
unknown).
We use a nonparametric method to perform a test of hypothesis.
Hypothesis Tests About µ : σ Known
Three Possible Cases
Prem Mann, Introductory Statistics, 9/E
Copyright © 2015 John Wiley & Sons. All rights reserved.
Two Procedures
Two procedures to make tests of hypothesis
1. The p-value approach- Under this procedure, we calculate
the p-value for our sample statistic and then we compare it
with a specific significance level. Then we make a decision.
P- value stands for Probability over here.
2. The critical-value approach- will be explained later
P value approach
Assuming that the null hypothesis is true, the p-value can be
defined as the probability that a sample statistic (such as the
sample mean) is at least as far away from the hypothesized value
in the direction of the alternative hypothesis as the one obtained
from the sample data under consideration. Note that the p–value
is the smallest significance level at which the null hypothesis is
rejected.
Prem Mann, Introductory Statistics, 9/E
Copyright © 2015 John Wiley & Sons. All rights reserved.
Figure 9.5 The p–Value for a Right-Tailed
Test
Figure 9.6 The p–Value for a Two-Tailed
Test
P value approach
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For a one-tailed test, the p-value is the area in the tail of the
sampling distribution curve beyond the observed sample
statistic.
For a left-tailed test the p-value will be the area in the left
tail, left of where the sample statistic lies.
Similarly for a right- tailed test the p-value will be the area in
the right tail, right of where the sample statistic lies.
For a 2-tail test, the p- value is twice the area in any tail of
the sampling distribution curve, beyond the observed value of
the sample statistic.
Calculating the z Value for 𝒙
When using the normal distribution, the value of z for
a test of hypothesis about μ is computed as follows:
z
x 
x
𝒙 for

where  x 
n
The value of z calculated for 𝒙 using this formula is also called
the observed value of z.
Steps to Perform a Test of Hypothesis Using the
p–Value Approach
1.
2.
3.
4.
State the null and alternative hypothesis.
Select the distribution to use.
Calculate the p–value.
Make a decision.
Example 9-1
At Canon Food Corporation, it takes an average of 90
minutes for workers to learn a food processing job.
Recently the company installed a new food processing
machine. The supervisor at the company wants to find if the
mean time taken by workers to learn the food processing
procedure on this new machine is different than 90
minutes. A sample of 20 workers showed that it took, on
average, 85 minutes for them to learn the food processing
procedure on the new machine. It is known that the
learning times for all workers are normally distributed
with a population standard deviation of 7 minutes.
Find the p–value for the test that the mean learning time
for the food processing procedure on the new machine is
different from 90 minutes. What will your conclusion be if α
= .01?
Example 9-1: Solution
Step 1: State the Null and Alternate Hypotheses
H0: μ = 90
H1: μ ≠ 90
Step 2: Select the distribution to use
The population standard deviation σ is known, the sample size
is small (n < 30), but the population distribution is normal. So
we will use the normal distribution to find the p–value and
make the test.
Example 9-1: Solution
Step 3: Calculate the p-value
The sign in the alternative hypothesis indicates that the
test is two-tailed. The p-value is equal to twice the area
in the tail of the sampling distribution curve of to the left
of 𝑥=85 as shown in the Figure.
To find this area, we first find the z value for as follows:

7
 1.56524758 minutes
n
20
x 
85  90
z

 3.19
x
1.56524758
x 

Example 9-1: Solution
The area to the left of 𝑥 =85 is equal to the area under the
standard normal curve to the left of z =-3.19. From the normal
distribution table, the area to the left of z=-3.19 is 0.0007.
The curve is symmetric. Hence this same area also lies in the
right tail
Consequently, the p-value is
p-value = 2(.0007) = 0.0014
Figure 9-7 The p-Value for a Two-Tailed Test
Example 9-1: Solution
Step 4: Make a Decision
When the area of the significance level is ≥, the area
denoted by the p value we reject the Null.
Because α = .01 is greater than the p-value of 0.0014, we
reject the null hypothesis at this significance level.
Therefore, we conclude that the mean time for learning the
food processing procedure on the new machine is different
from 90 minutes.
Example 9-2
The management of Priority Health Club claims that
its members lose an average of 10 pounds or
more within the first month after joining the club.
A consumer agency wanted to check whether this
claim is false.
They take a random sample of 36 members of this
health club and find that the members lost an
average of 9.2 pounds within the first month of
membership. The population standard deviation
is 2.4 pounds.
Find the p–value for this test. What will your
decision be if α = .01? What if α = .05?
Example 9-2: Solution
Step 1: State the Null and Alternate Hypotheses
H0: μ ≥ 10
H1: μ < 10
Step 2: Select the distribution to use
The population standard deviation σ is known, the sample
size is large (n > 30). Due to the Central Limit Theorem,
we will use the normal distribution to find the p–value and
perform the test.
Example 9-2: Solution
Step 3: Calculate the p-value
The < sign in the alternative hypothesis indicates that
the test is left-tailed. The p-value is given by the area to
the left of x=-9.2 under the sampling distribution curve
of as shown in Figure. To find this area, we first find the
z value for as follows:

2.4
x 

 .40
n
36
x   9.2  10
z

 2.00
x
.40
Example 9-2: Solution
The area to the left of 𝑥=-9.2 under the sampling distribution of
𝑥 is equal to the area under the standard normal curve to the left
of z=-2.00. From the normal distribution table, the area to the
left of z=-2.00 is .0228. Consequently,
p-value = 0.0228
Figure 9-8 The p-Value for a Left-Tailed Test
Example 9-2: Solution
Step 4: Make a Decision
When the area of the significance level is ≥ the area
denoted by the p value we reject the Null.
Since α = .01 is less than the p-value of .0228, we do not
reject the null hypothesis at this significance level.
Consequently, we conclude that the mean weight lost within
the first month of membership by the members of this club is
10 pounds or more.
Because α = .05 is greater than the p-value of .0228, we
reject the null hypothesis at this significance level.
Therefore, we conclude that the mean weight lost within the
first month of membership by the members of this club is
less than 10 pounds.