CHAPTER 6
STATISTICAL INFERENCE:
HYPOTHESIS TESTS
1.
2.
3.
4.
The Concept of Hypothesis Testing
The General Methodology of Hypothesis Testing
2.1. The Procedure
2.1.1. The Null Hypothesis versus the Alternative Hypothesis
2.1.2. The Type I Error versus Type II Error
2.1.3. Two-Tailed Hypothesis Tests versus One-Tailed Hypothesis Tests
2.1.3.1. Two-Tailed Tests
2.1.3.1.1. Decision Rules
2.1.3.1.2. The Relationship Between the Confidence Interval and the Acceptance Region for
the TOH
2.1.3.1.3. Type I and Type II Errors Revisited
2.1.3.2. One-Tailed Tests
2.1.3.2.1. Lower Tail Test
2.1.3.2.2. Upper Tail Test
2.1.4. How to Set Up the Null and Alternative Hypotheses
Hypothesis Test for the μ—Small Samples From Normal Populations
Test of Hypothesis on Population Proportion π
1. The Concept of Hypothesis Testing
In addition to the confidence interval, the hypothesis test is another approach to making inferences about a
population parameter using a sample statistic. To compare the two approaches, in the confidence interval we
have no prior knowledge, judgment or claim about the population parameter. We take a sample and build an
appropriate interval around the sample statistic, for a given level of confidence, to estimate the range of
values within which the population parameter may fall.
In the hypothesis test, in contrast, we have some prior knowledge, judgment or claim about the population
parameter. In other words, we have a hypothesis about the population parameter, which may be accepted or
rejected depending on the analysis of the sample data.
2. The General Methodology of Hypothesis Testing
To explain the theoretical foundation of hypothesis tests we will start with the formula for the confidence
interval for μ with large sample size 𝑛:
𝐿, 𝑈 = 𝑥̅ ± 𝑧𝛼⁄2 se(𝑥̅ )
where se(𝑥̅ ) = 𝑠⁄√𝑛. Note that the interval is built around 𝑥̅ , where the expression 𝑧𝛼⁄2 se(𝑥̅ ) is the familiar
marginal error—MOE. The 𝑀𝑂𝐸 plays a similar role in the hypothesis test, as will be seen.
Now, to explain the methodology for the hypothesis test, consider the following example:
Chapter 6—Hypothesis Tests
Page 1 of 22
Example 1
Casual observation of vehicle speed on a freeway indicates that most vehicles exceed the speed limit of 70
mph. Suppose we want to test the hypothesis that mean speed is 75 mph. Accordingly, a random sample of
𝑛 = 110 vehicles is secretly clocked, yielding the following data:
65
74
82
73
80
86
80
69
87
83
78
83
84
66
81
80
90
77
84
80
67
62
91
69
92
84
69
64
65
84
82
68
75
65
88
86
89
85
66
76
92
76
66
85
88
78
84
83
83
81
64
91
76
88
89
69
64
79
66
78
90
81
72
66
77
84
64
65
65
87
62
83
75
78
74
92
84
87
86
89
82
87
78
72
73
68
91
76
90
87
76
72
85
71
67
86
62
89
70
73
68
83
65
89
72
73
70
62
70
72
The mean speed obtained from this data is 𝑥̅ = 77.5 mph. Does this sample provide significant evidence that
the mean speed of all vehicles is different from 75 mph? Should we reject the hypothesis that the population
mean speed is 75 mph?
Note that 𝑥̅ = 77.5 is obtained from a single sample. Now that we have learned about the sampling
distribution of 𝑥̅ , we know that there are infinite number of samples of size 𝑛 = 110, each yielding a different
𝑥̅ values. These values are normally distributed with the population mean 𝜇 as their center of gravity.
Therefore, it is inevitable that the mean obtained from a single sample of size n will deviate from the
population mean. The question is then: is the deviation of the sample mean from the hypothesized
population mean significant? To answer this question, we must determine whether the deviation is due to
sampling error. That is, does 𝑥̅ = 77.5 mph fall within the margin of sampling error (𝑀𝑂𝐸) from the
hypothesized population mean? If the deviation is within 𝑀𝑂𝐸, then we can conclude that this is a "natural"
deviation, that 77.5 is one of the 𝑥̅ values that fall within the 95% interval in the sampling distribution.
Therefore, this 𝑥̅ belongs to the sampling distribution which has the center of gravity μ = 75 mph, the
hypothesized population mean.
If the sample 𝑥̅ value falls within the ±𝑀𝑂𝐸 from µ0 = 75 (the expression µ0 implies that this is the
hypothesized, rather than the actual, population mean), then the deviation of the sample value from the
hypothesized population mean is said to be not significant—the deviation is due to the sampling error. If,
however, the 𝑥̅ value falls outside the interval µ0 ± 𝑀𝑂𝐸, then the deviation is said to be significant—the
deviation is not due the sampling error. This 𝑥̅ belongs to a different sampling distribution with a center of
gravity other than the hypothesized µ0 . Then it can be argued that the hypothesized µ0 is not the true
population mean. We should reject the hypothesis that the population mean is equal to 75 mph.
What constitutes a significant deviation? How do you determine the acceptable 𝑀𝑂𝐸 for a hypothesis test?
This is the question that the test of hypothesis attempts to answer.
2.1. The Procedure
The main task in performing a test of hypothesis is to find the margin of sampling error, 𝑀𝑂𝐸. This would
provide us with the decision rule, the criterion, to determine whether to reject the hypothesis.
Chapter 6—Hypothesis Tests
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2.1.1.
The Null Hypothesis versus the Alternative Hypothesis
To obtain the decision rule, first you must state the claim (the hypothesis) about the population mean in a
prescribed way. The claim consists of two components, a null hypothesis, denoted by 𝑯𝟎 , and an
alternative hypothesis, 𝑯𝟏 . For the vehicle speed example, we state the null and alternative hypotheses as
follows.
The null hypothesis:
The alternative hypothesis:
𝐻0 : µ = 75
𝐻1 : µ ≠ 75
The null hypothesis states that the population mean is equal to is 75 mph; the alternative hypothesis states
that the population mean is not equal to, or different than, 75 mph.
2.1.2.
The Type I Error versus Type II Error
Once you state your hypothesis, you must deal with the following dilemma involving hypothesis tests. Since
the test of hypothesis involves the sampling distribution, in deriving a conclusion from the results of the test
that is based on a random sampling process, there is always a chance that you may arrive at a wrong
conclusion about the hypothesis. A wrong conclusion can happen in two ways.
1) Reject a true null hypothesis. This is called a Type I Error.
2) Not reject a false null hypothesis. This is called a Type II Error.
There is always a chance or probability that you may commit either one of the two errors. The probability of
committing a Type I error is denoted by α and that of committing a Type II error is denoted by β. Reducing α,
for a given sample size, comes only at the cost of increasing β.
Performing a test of hypothesis is very similar to conducting a trial in a criminal court. Given the evidence
that a crime is committed, the defendant or the accused is charged for or accused of committing the crime.
The purpose of the trial is to establish the defendant’s guilt or innocence. The null hypothesis is that the
defendant is innocent (the accused is presumed innocent) and the alternative is that he is guilty. If the jury
finds an innocent person guilty, it has rejected a true null hypothesis; it has, therefore, made a Type I error.
On the other hand, if the jury finds a guilty person not guilty, it has not rejected a false null hypothesis; it has,
therefore, made a Type II error. The following table shows the four possible situations resulting from a test of
hypothesis (or a court trial).
The null hypothesis 𝑯𝟎 (presumed innocent)
𝑯𝟎 is True
𝑯𝟎 is rejected
𝑯𝟎 is not
rejected
Incorrect decision (Type I Error)
The accused is innocent and he is found
guilty.
Probability = α
Correct decision (no error)
The accused is innocent and he is found
not guilty.
Probability = 1 − α
𝑯𝟎 is False
Correct decision (no error)
The accused is guilty and he is found
guilty.
Probability = 1 − β
Incorrect decision (Type II Error)
The accused is guilty and he is found
not guilty.
Probability = β
In the hypothesis test, the burden of proof is always on the alternative hypothesis. In a criminal court, the
burden of proof is on the prosecutor. The prosecutor must convince the jury, show beyond a reasonable
doubt, that the defendant is guilty. Therefore, we want to make it unlikely to reject the null hypothesis unless
the evidence is "very strong" or "significant". In a criminal court, “significant” means “beyond a reasonable
doubt”. We want to make it unlikely to find the defendant guilty unless guilt is established beyond a
Chapter 6—Hypothesis Tests
Page 3 of 22
reasonable doubt. For this reason the α, the probability of rejecting the null hypothesis, is always assigned a
small value—typically, 5 percent in statistical hypothesis tests. The α value is also called the level of
significance of the test.
Note that in a confidence interval, α is the percentage of all possible intervals built around sample means that
do not capture the population mean. That was because α% of sample means fall outside the margin of error
𝑀𝑂𝐸 = 𝑧𝛼⁄2 se(𝑥̅ ). In a test of hypothesis α plays a similar role. If the randomly selected sample yields an 𝑥̅
value which falls outside the prescribed margin of error, we would wrongly reject the null hypothesis. And
there is always an α% chance of doing that.
Since committing a Type I Error is considered as the more serious of the two errors (finding an innocent person
guilty), the threshold probability (the level of significance α) is set in advance. The probability of Type II
Error (β), however, varies based on several factors, one of the them being α.
2.1.3.
Two-Tailed Hypothesis Tests
To determine the acceptance region, like the confidence interval, we need a margin of (sampling) error. The
form of the 𝑀𝑂𝐸 in the hypothesis test depends on the null hypothesis to be rejected. If the null hypothesis is
that µ0 = 75 (the population mean is equal to 75), then using the margin of error
MOE = 𝑧𝛼⁄2 se(𝑥̅ )
the interval which would contain 1 − 𝛼 percent of all sample means would be
𝐿, 𝑈 = 𝜇0 ± 𝑧𝛼⁄2 se(𝑥̅ )
Here the hypothesis test is said to be a two-tailed test. The reason this is called a two-tailed test is that no
matter what the value of the sample statistic 𝑥̅ , whether it is greater than or less than the hypothesized mean,
there is always some evidence against the null hypothesis in terms of the difference between the value of the
sample statistic and value stated as the null hypothesis. The purpose of the test (the trial) is to gauge the
significance of the difference in either direction from the null mean. The significance of the difference will be
measured relative to the margin of error on either side. In the 𝑀𝑂𝐸 formula, therefore, you must use 𝑧𝛼⁄2.
The null and alternative hypotheses for a TWO-TAIL TEST
The null hypothesis:
The alternative hypothesis:
𝐻0 : µ = 𝜇0
𝐻1 : µ ≠ 𝜇0
The vehicle speed example is a two-tail test. Test the null hypothesis that the population mean vehicle speed
is equal to 75.
𝐻0 : µ = 75 mph
𝐻1 : µ ≠ 75 mph
We select α = 0.05 (allowing for 5% chance of committing a Type I error, that is, rejecting a true null
hypothesis). Going back to the sample data shown above, the sample mean and standard deviation are
obtained as: 𝑥̅ = 77.5 and 𝑠 = 8.94. To determine the margin of error, first compute the standard error of 𝑥̅ .
se(𝑥̅ ) = 8.94⁄√110 = 0.852
Given α = 0.05, the other component of MOE, 𝑧𝛼⁄2 , is 𝑧0.025 = 1.96. The margin of error is then
Chapter 6—Hypothesis Tests
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MOE = (1.96)(0.852) = 1.67
The interval is then
𝐿, 𝑈 = 75 ± 1.67 = [73.23,76.67]
Since 𝑥̅ = 77.5 falls outside this interval, we conclude that the deviation is significant, and reject 𝐻0 : µ = 75.
In the diagram below, the interval [73.23,76.67] is labeled as the “acceptance region”. The sample mean falls
outside this region.
x̅ = 77.5
73.33
µ₀ = 75
76.67
73.33 ≤ Acceptance Region ≤ 76.67
2.1.3.1.
The Relationship Between the Confidence Interval and the Acceptance
Region for the TOH
We can use the above diagram to observe how the confidence interval for µ and the acceptance region for a
two-tail test of hypothesis are related. The margin of error for a 95% confidence interval for the vehicle
speed example is:
MOE = 𝑧𝛼⁄2 se(𝑥̅ ) = 1.96 × 0.852 = 1.67
The lower and upper boundaries of the confidence interval are:
𝐿, 𝑈 = 𝑥̅ ± MOE = 77.5 ± 1.67 = (75.83,79.17)
Chapter 6—Hypothesis Tests
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73.33
µ₀ = 75
76.67
= 75.83
= 77.5
= 79.17
Note that this interval does not capture the null mean µ₀ = 75. You can thus use a confidence interval to
observe if the null mean 𝜇0 falls within the interval. If it does not, then you reject the null hypothesis.
2.1.3.2.
Type I and Type II Errors Revisited
In this example, we rejected 𝐻0 = 75 because the sample mean 𝑥̅ = 77.5 happened to fall outside the 𝑀𝑂𝐸.
That is, this sample mean was not one of the 95% of 𝑥̅ values that would fall within the interval 75 ± 1.67.
Now we can ask the question, “what if the population mean were in fact 75 mph?”. If that were the case, then
5% of 𝑥̅ values would fall outside the interval 75 ± 1.67. Therefore, if 𝑥̅ = 77.5 belonged to this 5%, then we
have rejected a true 𝐻0 ; we have made a Type I error.
Suppose now we take another sample of 𝑛 = 110 and obtain 𝑥̅ = 76.2 mph and 𝑠 = 8.66. Note that with
𝑀𝑂𝐸 = 1.96(8.66⁄√110) = 1.6 the boundaries of the acceptance region are:
𝑥̅𝐿 , 𝑥̅𝑈 = 75 ± 1.6 = (73.4,76.6)
Then 𝑥̅ = 76.2 falls inside the “acceptance region” under the 𝐻0 distribution. Therefore, we conclude that this
sample mean belongs to the 𝐻0 distribution and do not reject the null hypothesis 𝐻0 : µ = 75. But, what if the
population mean is some number other than 75? Suppose the true population mean speed is µ1 . This is the
center of gravity of the alternative sampling distribution represented in the following diagram by 𝐻1 , and 𝑥̅ =
76.2 belongs to that distribution. Thus, by wrongly concluding that 𝑥̅ = 76.2 belongs to the 𝐻0 distribution,
we have not rejected a false null hypothesis. We have, therefore, committed a Type II error.
Chapter 6—Hypothesis Tests
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73.4
µ₀ = 75
76.6
µ₁
The following is a graphic representation of the four scenarios involving a hypothesis test:
o
o
o
o
𝐻0 is true and is not rejected: No error.
𝐻0 is true and is rejected: Type I error.
𝐻0 is false and is not rejected: Type II error.
𝐻0 is false and is rejected: No error.
Chapter 6—Hypothesis Tests
µ₀ = 75
µ₀ = 75
µ₀ = 75
µ₀ = 75
Page 7 of 22
2.1.3.3.
Decision Rules for Rejecting 𝑯𝟎
The decision rule is always set up to reject the null hypothesis. There are two ways to set up the decision
rule. Both are derived from the 𝑀𝑂𝐸 formula. The role of 𝑀𝑂𝐸 here is that, if the null hypothesis 𝐻0 were
true, then 1 − 𝛼 percent of the sample means must fall within the 𝑀𝑂𝐸. Thus 𝑀𝑂𝐸 becomes the criterion for
rejecting 𝐻0 . We will reject 𝐻0 , that is, we conclude the deviation 𝑥̅ − 𝜇0 is significant, only when 𝑥̅ falls
outside the 𝑀𝑂𝐸, when the (absolute value of) deviation of 𝑥̅ from 𝜇0 exceeds 𝑀𝑂𝐸.
We reject 𝐻0 if,
𝑥̅ − 𝜇0 > 𝑀𝑂𝐸
Substituting for 𝑀𝑂𝐸, we have,
𝑥̅ − 𝜇0 > 𝑧𝛼⁄2 se(𝑥̅ )
Dividing both sides of in inequality by se(𝑥̅ ) gives us,
𝑥̅ − 𝜇0
> 𝑧𝛼⁄2
se(𝑥̅ )
The term on the left-hand-side above is called the test statistic (𝑻𝑺)and 𝑧𝛼⁄2 is the critical value (𝑪𝑽).
Decision Rule (a)—Reject H0 if
̅ − µ𝟎
𝒙
> 𝒛𝜶⁄𝟐
̅)
𝐬𝐞(𝒙
|𝑻𝑺| ≡ |𝒛| > 𝑪𝑽 ≡ 𝒛𝜶⁄𝟐
Note that the test statistic when you compute the test statistic
𝑥̅ −𝜇0
se(𝑥̅ )
, the result is the 𝑧 score. Also note the
absolute value lines around the test statistic. This means that when the test statistic is negative, to avoid the
confusion arising from the negative sign regarding the direction of the inequality, use the absolute value.
Now back to the vehicle speed example. The mean obtained from the sample is 𝑥̅ = 77.5. The objective of
this exercise is to see if the deviation of the sample mean and the hypothesized mean (𝑥̅ − µ0 = 2.50) is
significant. If this difference exceeds MOE, then the difference is significant and it will lead us to reject the
null hypothesis. The difference is:
𝑥̅ − µ0 = 77.5 − 75 = 2.5
Using se(𝑥̅ ) = 0.852, the test statistic is,
𝑧=
𝑥̅ − 𝜇0
2.5
=
= 2.93
se(𝑥̅ )
0.852
and the critical value is,
𝑧𝛼⁄2 = 𝑧0.025 = 1.96
Chapter 6—Hypothesis Tests
Page 8 of 22
𝑇𝑆 = 2.93 > 𝐶𝑉 = 1.96, then the deviation is significant. Therefore, we reject the null hypothesis that µ0 =
75. We conclude that the population mean vehicle speed is different from 75 mph.
The alternative approach for determining if the difference 𝑥̅ − 𝜇0 is significant is to find the tail area
associated with the value of the test statistic. That is, find P(𝑧 > 𝑇𝑆). Using the z table, this tail area is:
P(𝑧 ≥ 𝑇𝑆) = P(𝑧 ≥ 2.93) = 0.0017
When the test is a two-tail test, double the computed probability (2 × 0.0017 = 0.0034) and compare it to
α = 0.05. Note that 0.0034 is now the computed probability of Type I error. With 𝑝𝑟𝑜𝑏 𝑣𝑎𝑙𝑢𝑒 = 0.0034, there
is about 0.58% probability that we might reject a true null hypothesis. Since we are allowing 5% as the
“comfort zone” or threshold probability for rejecting a true null, the computed probability 0.0034 is clearly
within this comfort zone. There is only a 0.34% chance that we will be rejecting a true null, or committing a
Type I error. This approach to the hypothesis test is the probability (𝑝𝑟𝑜𝑏) value approach.
Decision Rule (b)—Reject H0 if
𝟐 × 𝐏(𝒛 > 𝑻𝑺) < 𝜶
p-value < level of significance
For a two-tail test, in Decision Rule (b) the 𝒑𝒓𝒐𝒃 value is twice the tail area corresponding to TS. If the
𝒑­value < α, then reject 𝑯𝟎 .
Summary of Steps For a Two-Tail Tests
a.
State the null and alternative hypotheses.
𝐻0 : µ = µ0
𝐻1 : µ ≠ µ0
a.
Specify the level of significance α.
b.
Use any of the two methods to reject or not reject the null hypothesis
Decision Rule (a)—Test Statistic
𝑥̅ − 𝜇0
se(𝑥̅ )
i.
Compute the test statistic
𝑇𝑆 =
ii.
Determine the critical value
𝐶𝑉 = 𝑧α⁄2
iii.
Reject H0 if
𝑇𝑆 > 𝐶𝑉
Chapter 6—Hypothesis Tests
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Decision Rule (b)—The 𝒑𝒓𝒐𝒃 value
𝑥̅ − 𝜇0
se(𝑥̅ )
i.
Compute the test statistic
𝑇𝑆 =
ii.
Find the probability value
2 × 𝑃(𝑧 > 𝑇𝑆)
iii.
Reject H0 if
p-value < α
(The two tail areas for TS)
2.1.4. One-Tailed Tests
In many cases the null hypothesis is that the population mean is either at least (greater than or equal to), or is
at most (less than or equal to) some value. In these cases the sample statistic 𝑥̅ may contradict the null
hypothesis in only one direction. For example, if 𝐻0 is 𝜇 ≥ 75, the test is of interest only if 𝑥̅ is less than 75.
Only this way does the sample statistic contradict the null hypothesis and we want to test whether this is a
significant contradiction. If the sample mean turns out to be greater than 75, then it confirms the null and,
therefore, there is no need for the test.1 This is why the significance of the deviation will be measured relative
to the margin of error only in one direction. Regarding the level of significance α, to maintain the same
probability of rejecting a true null hypothesis as in a two-tailed test, the whole α must be used. Thus, in the
𝑀𝑂𝐸 formula we use 𝑧α instead of 𝑧α⁄2 .
𝑀𝑂𝐸 = 𝑧𝛼 𝑠𝑒(𝑥̅ )
Here the hypothesis test is said to be a one-tail test.
In the above example, we conducted a two-tail test, testing the null hypothesis 𝐻0 : µ = 75 mph against the
alternative 𝐻1 : µ ≠ 75 mph. What if the concern was the mean vehicle speed is 75 mph or more (at least 75
mph). In this case we would be conducting a one-tail test, testing the null hypothesis 𝐻0 : µ ≥ 75 mph against
the alternative 𝐻1 : µ < 75 mph.
2.1.4.1.
Lower Tail Test
If the sample of 110 vehicles yields a mean which is less than 75 mph, this may be evidence that the
population mean speed is less than 75 mph. The question is, however, is the evidence conclusive? Is the
sample mean significantly less than 75? How far should the sample mean fall below the null mean, 𝜇0 ≥ 75,
before we conclude that the evidence is significant?
To set up the test, the null hypothesis 𝐻0 must be that the mean is equal to or greater than 75 mph. To reject
this hypothesis, the sample evidence must be significant. That is, the sample mean must be significantly less
than 75 mph. Generally, for a lower-tail test, the null and alternative hypotheses are written as
The null and alternative hypotheses for a LOWER-TAIL TEST
The null hypothesis:
The alternative hypothesis:
1
𝐻0 : µ ≥ 𝜇0
𝐻1 : µ < 𝜇0
If there is no evidence the “defendant” has committed the crime, then there would be no trial.
Chapter 6—Hypothesis Tests
Page 10 of 22
For this example, the null and alternative hypotheses are written as:
𝐻𝑜 : µ ≥ 75
𝐻1 : µ < 75
This is a lower-tail test, as indicated by "<" (a strict inequality) in the alternative hypothesis.
Example 2
To perform a test, let us continue with the example of clocking a random sample of 𝑛 = 110 vehicles.
Suppose the sample yields 𝑥̅ = 73.9 mph and 𝑠 = 8.82. Note that 𝑥̅ = 73.9 < µ0 = 75 implies that the
evidence from the sample contradicts 𝐻0 . The question is, is this a significant evidence contradicting the null
hypothesis?
Let us compute the 𝑀𝑂𝐸, determine the acceptance region for the test, and see where 𝑥̅ falls relative to the
acceptance region.
se(𝑥̅ ) =
8.82
√110
= 0.841
𝑀𝑂𝐸 = 𝑧𝛼 se(𝑥̅ ) = 1.64 × 0.841 = 1.38
The following diagram shows the acceptance region and where the 𝑥̅ value falls. Note that the acceptance
region is now bounded only on the left tail.
𝑥𝐿 = µ0 − MOE = 75 − 1.38 = 73.62
x̅ = 73.90
73.62
µ₀ = 75
The sample statistic 𝑥̅ = 73.90 falls inside the acceptance region bounded on the left by 𝑥̅𝐿 = 73.62 mph. We
do not reject the null hypothesis and conclude that the population mean is not less than 75 mph.
The following show the decision rules for a lower tail test.
Chapter 6—Hypothesis Tests
Page 11 of 22
Decision Rule (a) for a Lower Tail Test—Reject H0 if
̅ − µ𝟎
𝒙
> 𝒛𝜶
̅)
𝐬𝐞(𝒙
|𝑻𝑺| ≡ |𝒛| > 𝑪𝑽 ≡ 𝒛𝜶
For our example,
𝑇𝑆 =
𝑥̅ − 𝜇0 73.9 − 75
=
= −1.31
se(𝑥̅ )
0.841
To avoid the confusion arising with the negative sign of the test statistic, use the absolute value of 𝑇𝑆 to
compare to the 𝐶𝑉. Thus,
|𝑇𝑆| = 1.31 < 𝐶𝑉 = 𝑧0.05 = 1.64
We do not reject the null hypothesis and conclude that the population mean speed is not less than 75 mph.
Decision Rule (b)—Reject H0 if
𝐏(𝒛 < 𝑻𝑺) < 𝜶
p-value < level of significance
To find the p-value
P(𝑧 < −1.31) = 0.0951
Since this is a one-tail test, we do not double the tail area. Thus,
𝑝­𝑣𝑎𝑙𝑢𝑒 = 0.0951 > 𝛼 = 0.05
We do not reject the null hypothesis.
i. Find the critical value.
ii. Find the test statistic:
iii. Reject H0 if −𝑇𝑆 < −𝐶𝑉, or |𝑇𝑆| > 𝐶𝑉
2.1.4.2.
𝐶𝑉 = 𝑧α = 𝑧0.05 = 1.64
𝑇𝑆 = (𝑥̅ − µ0 )⁄se(𝑥̅ ) = −1.10⁄0.841 = −1.31
Do not reject H0 since
𝑇𝑆 = −1.25 > −𝐶𝑉 = −1.64 𝑜𝑟 |𝑇𝑆| = 1.25 < 𝐶𝑉 = 1.64
Upper Tail Test
The upper-tail test applies when we want to test if the sample evidence is significantly greater than the value
stated in the null hypothesis for the population mean.
Example 3
A random sample n = 115 reimbursements for office visits to physicians paid by Medicare yielded a sample
mean of 𝑥̅ = $104.9 and a standard deviation of 𝑠 = $25.30. Does the sample provide significant evidence
Chapter 6—Hypothesis Tests
Page 12 of 22
that the mean reimbursement is greater than $100? Perform the test of hypothesis at a 5% level of
significance.
Since we want to determine if the sample mean 𝑥̅ = $104.9 is significantly greater than (>) $100, the null
hypothesis should be µ0 ≤ $100. Therefore, we must write the null and alternative hypotheses as:
𝐻0 : µ ≤ $100
𝐻1 : µ > $100
Let us compute the 𝑀𝑂𝐸, determine the acceptance region for the test, and see where 𝑥̅ falls relative to the
acceptance region.
se(𝑥̅ ) =
25.3
√115
= 2.36
𝑀𝑂𝐸 = 𝑧𝛼 se(𝑥̅ ) = 1.64 × 2.36 = 3.87
The following diagram shows the acceptance region and where the 𝑥̅ value falls. Note that the acceptance
region is now bounded only on the right tail.
𝑥𝑈 = µ0 + MOE = 100 + 3.87 = 103.87
x̅ = 104.90
µ₀ = 100
103.87
The sample mean 𝑥̅ = $104.9 falls outside the acceptance region. Hence, we reject the null hypothesis,
𝐻0 : µ ≤ $100, and conclude that the mean reimbursement is greater than $100.
Now let’s use the test statistic and p-value decision rules.
Decision Rule: Reject 𝑯𝟎 , if 𝑻𝑺 > 𝑪𝑽
𝑇𝑆 = 𝑧 =
𝑥̅ − 𝜇0 104.9 − 100
=
= 2.08
se(𝑥̅ )
2.36
𝐶𝑉 = 𝑧𝛼 = 𝑧0.05 = 1.64
Since 𝑇𝑆 = 2.08 > 𝐶𝑉 = 1.64, reject 𝐻0
Decision Rule: Reject 𝑯𝟎 , if p-𝐯𝐚𝐥𝐮𝐞 < 𝜶
p-value = P(𝑧 > 𝑇𝑆) = P(𝑧 > 2.08) = 0.0188
Chapter 6—Hypothesis Tests
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Reject H0 since p-value = 0.0188 < 𝛼 = 0.05
Remark:
What would the conclusion be if α = 0.01?
Since 𝑇𝑆 = 2.08 < 𝐶𝑉 = 𝑧0.01 = 2.33, do not reject 𝐻0 .
Since p-value = 0.0188 < 𝛼 = 0.01, do not reject 𝐻0 .
𝑀𝑂𝐸 = 𝑧α se(𝑥̅ ) = 2.33(8.82⁄√115) = 5.56
Example 4
A light bulb manufacturer claims the mean life of its light bulbs is at least 1,000 hours. To perform a test of
hypothesis at 5 percent level of significance, a sample of 𝑛 = 105 light bulbs yields an average life of 989.2
and a standard deviation of 56 hours. Should the manufacturer's claim be rejected? Use 𝛼 = 0.05.
The problem is asking if we should reject the manufacturer’s claim that the mean life is at least 1,000. At
least means “no less than” or “greater than or equal to”, the symbol for which is “≥”. This is the null
hypothesis symbol. The alternative is “<”.
𝐻0 : 𝜇 ≥ 1,000
𝐻1 : 𝜇 < 1,000
𝑠 = 56
se(𝑥̅ ) = 56⁄√105 = 5.465
𝛼 = 0.05
𝑧𝛼 = 𝑧0.05 = 1.64
Decision Rule (a)—Reject H₀ if |𝑻𝑺| < 𝑪𝑽
𝑛 = 49
i. Find the critical value:
ii. Find the test statistic:
iii. Reject H0 if |𝑇𝑆| > 𝐶𝑉
𝐶𝑉 = 𝑧α = 𝑧0.05 = 1.64
|𝑇𝑆| = (𝑥̅ − µ0 )⁄se(𝑥̅ ) = −10.8⁄5.465 = 1.98
Reject H0 since |𝑇𝑆| = 1.98 > 𝐶𝑉 = 1.64
Decision Rule (b)—Reject H₀ if p-value < 𝜶
i. Find the test statistic:
ii. Find the 𝑝𝑟𝑜𝑏 value: P(𝑧 < 𝑇𝑆)
iii. Reject H0 if 𝑝𝑟𝑜𝑏 𝑣𝑎𝑙𝑢𝑒 < 𝛼
𝑇𝑆 = (𝑥̅ − µ0 )⁄se(𝑥̅ ) = −10.8⁄5.465 = −1.98
P(𝑧 < −1.98) = 0.0239
Reject 𝐻0 since p-value = 0.0239 < 𝛼 = 0.05
Reject the manufacturer’s claim that the mean life is at least 1,000 hours and conclude that it is less the 1,000
hours.
2.1.5. How to Set Up the Null and Alternative Hypotheses
The most important part of performing a hypothesis test is stating the correct null and alternative
hypotheses. The incorrect statement of the hypotheses will invariably lead you to a wrong conclusion about
the test. If you are confused about setting up the hypotheses, hopefully the following guidelines will help.

Never put the equal sign in the alternative hypothesis. The following symbols should not appear
in the alternative hypothesis: "=", "≤", "≥". These symbols belong to the null hypothesis. Depending
on the nature of the test, the alternative hypothesis may contain any of the following: "≠", ">", "<".
Chapter 6—Hypothesis Tests
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
Following the above directions, after you state your null and alternative hypotheses, make certain
that the sample evidence contradicts the null hypothesis (and agrees with the alternative).
Remember, the reason we conduct a hypothesis test is to determine if the sample evidence is
significant in order to reject the null. In a two tail test, the sample evidence will always be different,
or contradict, the null. So, there is no confusion. However, in a one tail test the reason we conduct
the test is that there is evidence against the null, and we want to determine if the evidence is
significant. For example, if in a problem you set your null and alternative hypothesis as, say,
𝐻0 : 𝜇 ≥ 100
𝐻1 : 𝜇 < 100
and the sample evidence is
𝑥̅ = $110,
then the sample evidence does not contradict the null. There is no evidence that the population mean
is less than $100; there is no evidence to reject the null. This should be a warning that your
hypotheses statement is incorrect. The correct statement should be,
𝐻₀: 𝜇 ≤ $100
𝐻₁: 𝜇 > $100
Now the sample evidence, 𝑥̅ = $110, contradicts the null. There is evidence the mean is greater than
$100, but you want to determine if 𝑥̅ is significantly greater than 100 in order for you to reject the
null.

Generally, any hypothesis test which involves challenging the status quo, the prevailing practice or
belief, the challenger's viewpoint should be the alternative hypothesis. If you want to prove the
prevailing practice or belief wrong, you have to provide significant proof, a proof which is "beyond a
reasonable doubt". Consider the following examples
o The production team of a manufacturing company has designed a new production process
which is supposed to lower the average production cost. To implement the new process, the
production team must convince the management that the average cost is lower with the
proposed process than the current process. Suppose the current average cost is $10. The
production team must provide significant evidence that the average cost under their
proposed process is lower. Therefore, the null and alternative hypotheses must be stated as:
𝐻₀: 𝜇 ≥ $10
𝐻₁: 𝜇 < $10
Note that the null hypothesis states that the average cost is "no less than" $10. The task of the
production team is to show significant evidence to reject the null.
o
A pharmaceutical company has developed new drug to treat a certain type of cancer. Suppose 60% of
patients who take the existing drug experience remission. To prove that the new drug is more effective
than the current treatment, the company must convince, must provide significant evidence to the
medical community that the new drug is better, that the remission rate is higher. The null hypothesis,
to be rejected, then must be the new drug is no better:
𝐻₀: 𝜋 ≤ 60%
𝐻₁: 𝜋 > 60%
An interesting point to keep in mind in this example is that the medical community would require a
smaller level of significance for the test, say, 𝛼 = 0.01, compared to the typical 𝛼 = 0.05. This is to
reduce the probability of Type I error, to lower the likelihood of rejecting the "no better" hypothesis,
when it may be true.
Chapter 6—Hypothesis Tests
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
Another issue you should keep in mind in choosing the null and alternative hypothesis is: choose 𝐻0
such that, if the hypothesis is true, the consequence of rejecting it is costly, dire, etc...
In the problems that you deal with in this course, you should mainly be concerned about how the problem is
stated. The problem may be stated as the null hypothesis or the alternative hypothesis. For example, if you
are asked to test the hypothesis that the mean is "at least", say, $50, then you should recognize this as a null
statement: 𝐻₀: 𝜇 ≥ 0. The same problem may be state as: test the hypothesis that the mean is "less than"
$50. This is an alternative hypothesis statement: 𝐻₀: 𝜇 < 0. Just be careful to use the appropriate symbol
corresponding to the statement of the hypothesis. Then make sure that the equality sign in any form, "=",
"≥", or "≤", does not appear in the alternative hypothesis statement.
3. Hypothesis Test for the μ—Small Samples From Normal Populations
Like for confidence intervals, when the standard deviation of the population is unknown, the test of
hypothesis will use the t distribution, if the sample size is small.
Example 5
A filling machine fills bottles with a target mean of 12 ounces of beer. To test whether the target mean is
being achieved, a random sample of 20 bottles is selected with the following results (in ounces):
12.06
11.86
11.84
11.98
12.00
11.96
11.83
11.95
12.03
11.82
11.91
11.75
11.96
11.95
11.86
11.97
11.85
11.92
11.89
12.02
Perform the test at 5% level of significance.
Achieving the target mean implies that our null hypothesis should be 𝜇 = 12, and we want to find out if the
sample mean deviated from the target mean significantly.
𝐻0 : 𝜇 = 12
𝐻1 : 𝜇 ≠ 12
First we must compute the sample mean 𝑥̅ and the sample standard deviation 𝑠 from the sample:
𝑥̅ =
∑𝑥
= 12.032
𝑛
Find the standard error:
Find the t score:
𝑠=√
∑(𝑥 − 𝑥̅ )2
= 0.100
𝑛−1
se(𝑥̅ ) = 0.100⁄√20 = 0.022
𝑡𝛼/2,(𝑛−1) = 𝑡0.025,(19) = 2.093
Note that the margin of error is now computed using the 𝑡 distribution.
𝑀𝑂𝐸 = 𝑡𝛼/2,(𝑛−1) se(𝑥̅ ) = 2.093 × 0.022 = 0.05
The acceptance region is 𝐿, 𝑈 = 𝐻0 ± 𝑀𝑂𝐸
𝐿 = 12 − 0.05 = 11.05
𝑈 = 12 + 0.05 = 12.05
Chapter 6—Hypothesis Tests
Page 16 of 22
x̅ = 12.032
μ₀ = 12
11.95
x̅
12.05
11.95 ≤ Acceptance Region ≤ 12.05
Decision Rule (a)—Reject H₀ if 𝑻𝑺 > 𝑪𝑽
i. Find the critical value in terms of t
ii. Find the test statistic:
iii. Reject 𝐻0 if 𝑇𝑆 > 𝐶𝑉
𝐶𝑉 = 𝑡𝛼/2,(𝑛−1) = 𝑡0.025,(19) = 2.093
𝑇𝑆 = (𝑥̅ − 𝜇0 )⁄se(𝑥̅) = 1.431
Do not reject 𝐻0 since 1.431 < 2.093
Decision Rule (b)—Reject 𝑯𝟎 if probability value < 𝜶
i.
ii.
Find the test statistic
Find 2 × 𝑃(𝑡 > 𝑇𝑆)
𝑇𝑆 = (𝑥̅ − 𝜇0 )⁄se(𝑥̅) = 1.431
2 × 𝑃(𝑡 > 1.431) = 0.1687 2
(See the footnote and the note below)
Do not reject 𝐻0 since 0.1687 > 0.05.
iii. Reject 𝐻0 if p value < 𝛼.
IMPORTANT NOTE
Note that to compute P(𝑡 > 1.43) you must use a computer program that finds the tail area under the t curve
for a given t score and degrees of freedom. There are no tables to determine such areas (probabilities).
However, you can estimate this probability 𝑜𝑟𝑑𝑖𝑛𝑎𝑙𝑙𝑦 using the 𝑡 table as shown below:
df
16
17
18
19
20
21
22
23
0.100
1.337
1.333
1.330
1.328
1.325
1.323
1.321
1.319
0.050
1.746
1.740
1.734
1.729
1.725
1.721
1.717
1.714
0.025
2.120
2.110
2.101
2.093
2.086
2.080
2.074
2.069
0.010
2.583
2.567
2.552
2.539
2.528
2.518
2.508
2.500
0.005
2.921
2.898
2.878
2.861
2.845
2.831
2.819
2.807
You can still correctly guess, from a given 𝑡 value, whether the 𝑝𝑟𝑜𝑏 value is greater or less than a given level
of significance α (for a one-tail test) or 𝛼⁄2 (for a two-tail test). In the last example 𝑡 = 1.431. Given 𝑑𝑓 = 19,
the 𝑡 score increases as the tail area in the top row decreases (as we move to the right in the table). In the
above example, 𝑡 = 1.431 is greater than 1.328, the smallest 𝑡 score shown in the table associated with 𝑑𝑓 =
19. This means that the tail area associated with t score of 1.431 must be greater than 0.100. So, the
2
Here the Excel command =T. DIST. 2T(x, deg_freedom) is used, where =T. DIST. 2T(1.431,19) = 0.1687.
Chapter 6—Hypothesis Tests
Page 17 of 22
combined tail areas is definitely greater than the level of significance α. Therefore, we do not reject the null
hypothesis.
Example 6
A light bulb manufacturer claims the average life of its light bulbs is at least 1,000 hours. To perform a test of
hypothesis at 5 percent level of significance, a sample of 25 light bulbs yields an average life of 992.6 hours
with a sample standard deviation 𝑠 = 49.3 hours. Should the manufacturer's claim be rejected?
𝐻0 : 𝜇 ≥ 1,000
𝐻1 : 𝜇 < 1,000
Note that this is a lower tail test because 𝑥̅ − 𝜇0 = 992.6 − 1000 = −7.40 < 0.
𝑛 = 25
𝑥̅ = 992.6
se(𝑥̅ ) = 49.3⁄√25 = 9.86
𝑠 = 49.3
𝛼 = 0.05
Find the t score: 𝑇𝑆 = 𝑡𝛼,(𝑛−1) = 𝑡0.05,(24) = 1.711. [Here you must use 𝑡𝛼,(𝑛−1) , rather than 𝑡𝛼⁄2,(𝑛−1) , because
you are performing a one-tail test.]
Decision Rule (b)—Reject H₀ if |𝑻𝑺| > 𝑪𝑽
i. Find the critical value in terms of t
ii. Find the test statistic:
iii. Reject 𝐻0 if |𝑇𝑆| > 𝐶𝑉
𝐶𝑉 = 𝑡𝛼,(𝑛−1) = 𝑡0.05,(24) = 1.711
|𝑇𝑆| = |𝑥̅ − 𝜇0 |⁄se(𝑥̅) = 0.751
Do not reject 𝐻0 since 0.751 < 1.711
Decision Rule (c)—Reject 𝑯𝟎 if probability value < 𝜶
i. Find the test statistic
𝑇𝑆 = (𝑥̅ − 𝜇0 )⁄se(𝑥̅) = −0.751
ii. Find 𝑃(𝑡 < 𝑇𝑆)
𝑃(𝑡 < −0.751) = 0.2300 (See the note below)
iii. Reject 𝐻0 if p-value < 𝛼.
Do not reject 𝐻0 since 0.2300 > 0.05.
NOTE: Using Excel, =T.DIST.RT(0.751,24) = 0.2300. If a computer is not available, you can use the t table to
determine if the p-value is greater than or less than the level of significance:
df
23
24
25
0.100
1.319
1.318
1.316
0.050
1.714
1.711
1.708
0.025
2.069
2.064
2.060
0.010
2.500
2.492
2.485
0.005
2.807
2.797
2.787
Note that |t| = 0.751 is less than 1.318, the smallest of the shown t scores corresponding to df = 24, which is
associated with a tail area of 0.10, the largest of the shown tail areas. Thus, |t| = 0.751 must be associated
with a much larger tail area than 0.10, which, in turn, would exceed α = 0.05.
4. Test of Hypothesis on Population Proportion 𝝅
The hypothesis test about the population proportion follows a pattern similar to that for the population
mean. You compare the sample proportion 𝑝̅ to the value stated in the null hypothesis regarding 𝜋. If the
difference between 𝑝̅ and 𝜋0 exceeds MOE, then this difference is statistically significant and you reject the
null hypothesis.
Chapter 6—Hypothesis Tests
Page 18 of 22
Example 7
To test the hypothesis that the proportion of all Hoosier adults in the labor force with a 4-year college degree
is 26 percent, a sample of 600 Hoosier adults in the labor force is selected. The sample proportion is 27.3
percent. Test the hypothesis at 5 percent level of significance.
𝐻0 : 𝜋 = 0.26
𝐻1 : 𝜋 ≠ 0.26
This is a two-tail test, because we are testing the hypothesis that the population proportion is 26 percent.
𝑛 = 600
𝛼 = 0.05
𝑝̅ = 0.273
𝑧𝛼/2 = 1.96
To perform the test you need to determine the standard error of 𝑝̅. Use the following formula:
π0 (1 − π0 )
se(𝑝) = √
𝑛
Note that to find se(𝑝̅), unlike the standard error in the confidence interval problems, instead of the sample
proportion 𝑝̅ you use π0 in the formula. This is logical because we are presuming the population proportion
is the value specified in the null hypothesis.
0.26(1 − 0.26)
se(𝑝) = √
= 0.0179
600
The margin of error and the acceptance region for the test are determined as follows:
𝑀𝑂𝐸 = 𝑧𝛼⁄2 se(𝑝̅) = 1.96 × 0.0179 = 0.035
𝐿, 𝑈 = 𝜋0 ± 𝑀𝑂𝐸
𝐿 = 0.26 − 0.035 = 0.225
𝑈 = 0.26 + 0.035 = 0.295
p̅ = 0.273
0.225
π₀ = 0.26
0.295
0.225 ≤ Acceptance Region ≤ 0.295
The sample p̅ = 0.273 falls within the acceptance region.
Chapter 6—Hypothesis Tests
Page 19 of 22
Decision Rule (a)—Reject 𝑯𝟎 if 𝑻𝑺 > 𝑪𝑽
i. Find the critical value
ii. Find the test statistic 𝑇𝑆
iii. Reject H0, if 𝑇𝑆 > 𝐶𝑉
𝐶𝑉 = 𝑧𝛼⁄2 = 𝑧0.025 = 1.96
𝑇𝑆 = (𝑝̅ − 𝜋0 )⁄se(𝑝̅) = (0.273 − 0.26)⁄0.0179 = 0.73
Do not reject 𝐻0 since 0.73 < 1.96
Decision Rule (b)—Reject H₀ if probability value < α
i. Find 𝑇𝑆
𝑇𝑆 = 0.73
ii. Find the 𝑝𝑟𝑜𝑏 𝑣𝑎𝑙𝑢𝑒 2 × P(𝑧 > 𝑇𝑆) 2 × 𝑃(𝑧 > 0.73) = 2 × 0.2327 = 0.4654
iii. Reject 𝐻0 , if 𝑝𝑟𝑜𝑏 𝑣𝑎𝑙𝑢𝑒 < 𝛼:
Do not reject since 0.4654 > 0.05
The test of hypothesis provides that we should not reject the null hypothesis that 𝐻0 : 𝜋 = 0.26. Therefore we
conclude that the proportion of all Hoosier adults in the labor force with a 4-year college degree is 26 percent.
Example 8
A pest control company claims that no more than 15% of its customers need repeated treatment after a 90day warranty period. To test the validity of this claim, a consumer organization selected a sample of 300
customers and found that 57 needed repeated treatment after the 90-day warranty period. Is there evidence,
at 5% level of significance, that the claims is not valid?
Here the claim is "no more than" 15%... The symbol for "no more than" or "at most" is ≤. This symbol must
be stated in the null hypothesis. The alternative is then "greater than" 15%, which is shown as > 15%. This
makes the test an upper tail test.
𝐻0 : 𝜋 ≤ 0.15
𝐻1 : 𝜋 > 0.15
𝑛 = 200
𝑝̅ = 57⁄200 = 0.19
𝛼 = 0.05
𝑧𝛼 = 1.64
0.15(1 − 0.15)
se(𝑝) = √
= 0.0206
300
Compute 𝑀𝑂𝐸. Note that since this is a one tail test. Therefore, you must use 𝑧𝛼 , rather than 𝑧𝛼⁄2 , to obtain
𝑀𝑂𝐸.
𝑀𝑂𝐸 = 𝑧𝛼 se(𝑝̅ ) = 1.64 × 0.0206 = 0.034
For the acceptance region:
𝑈 = 𝜋0 + 𝑀𝑂𝐸 = 0.15 + 0.034 = 0.184
Chapter 6—Hypothesis Tests
Page 20 of 22
p̅ = 0.19
π₀ = 0.15
0.184
Acceptance Region ≤ 0.184
The sample statistic 𝑝̅ = 0.19 falls inside the acceptance region. Therefore, do not reject H0.
Decision Rule (a)—Reject H₀ if 𝑻𝑺 > 𝑪𝑽
i. Find the critical value
ii. Find the test statistic
iii. Reject H0, if 𝑇𝑆 > 𝐶𝑉.
𝐶𝑉 = 𝑧𝛼 = 𝑧0.05 = 1.64
𝑇𝑆 = (𝑝̅ − 𝜋0 )⁄se(𝑝̅) = 1.94
Reject 𝐻0 since 𝑇𝑆 = 1.94 > 𝐶𝑉 = 1.64
Decision Rule (b)—Reject H₀ if probability value < α
i. Find the test statistic
ii. Find the 𝑝𝑟𝑜𝑏 𝑣𝑎𝑙𝑢𝑒
iii. Reject 𝐻0 , if 𝑝𝑟𝑜𝑏 𝑣𝑎𝑙𝑢𝑒 < 𝛼
𝑇𝑆 = (𝑝̅ − 𝜋0 )⁄se(𝑝̅) = 1.94
P(z > 1.94) = 0.0.0262
Reject 𝐻0 since 𝑝𝑟𝑜𝑏 𝑣𝑎𝑙𝑢𝑒 = 0.0262 < 𝛼 = 0.05
Both methods indicate that the null hypothesis H0: π ≤ 0.15 should be rejected. The test does not support the
company’s claim that no more than 15% of its customers need repeated treatment after a 90-day warranty
period.
Example 9
To test the hypothesis that less than 40% of drivers on a certain highway obey the legal speed limit, in a
random sample of 700 vehicles clocked secretly, 252 observed the legal speed limit. Is there significant
evidence that less than 40% of drivers observe the legal speed limit? Perform the test at a 5 percent level of
significance.
Since the hypothesis to be tested is “less than” 40 percent (𝜋0 < 0.40), then this is a lower tail test:
𝐻0 : 𝜋 ≥ 0.40
𝐻1 : 𝜋 < 0.40
Compute the sample proportion:
𝑝̅ = 𝑥 ⁄𝑛 = 252⁄700 = 0.36
Since this is a lower tail test, the deviation of the sample proportion from the null value for the proportion
should be a negative value.
𝑝̅ − 𝜋0 = 0.36 − 0.40 = −0.04
se(𝑝̅ ) = √𝜋0 (1 − 𝜋0 )⁄𝑛 = √0.40(1 − 0.40)⁄700 = 0.0185
Chapter 6—Hypothesis Tests
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𝑀𝑂𝐸 = 1.64 × 0.0185 = 0.03
p̅ = 0.36
0.37
π₀ = 0.15
The test statistic is then,
𝑇𝑆 = (𝑝̅ − 𝜋0 )⁄se(𝑝̅)
|𝑇𝑆| = |0.36 − 0.40|/0.0185 = 2.16
At 𝛼 = 0.05, the critical value is,
𝐶𝑉 = 𝑧0.05 = 1.64
Decision rule: reject 𝐻0 , if |𝑇𝑆| > 𝐶𝑉:
Since |𝑇𝑆| = 2.16 > 𝐶𝑉 = 1.64, reject 𝐻0 . Conclude that less than 40 percent of drivers observe the legal
speed limit.
The p‐value for the test is
𝑃(𝑧 < −2.16) = 0.0154
Since 𝑝­ 𝑣𝑎𝑙𝑢𝑒 = 0.0154 < 𝛼 = 0.05, reject 𝐻0 .
Chapter 6—Hypothesis Tests
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