Chapter 7 – Hypothesis Testing with One Sample
7.1 Statistical Hypotheses
A statistical hypothesis is a mathematical claim about a population parameter.
Examples:

The mean height of women is less than 65 inches tall.  
65
,

The percentage of Floridians favoring a bullet train is 57%. p

.57

The average distance driven per year by Americans is more than 10,000 miles.
 
10, 000

At least 5% of Americans earn more than $100,000 per year. p

.05
Hypothesis Testing - Basic Procedure
If we wanted to know whether any of the above hypotheses are true, we would conduct a hypothesis test . When we test a statistical hypothesis, we follow the following basic procedure:
1.
Draw a
for the random variable in question.
2.
Determine if the results from the sample data are consistent or
.
3.
If the sample data is
o
a
. If the data is not significantly different, we would not reject the null hypothesis – there would not be any evidence to support the alternative hypothesis.
We can only “support” a claim if the null is rejected and we support the alternative.
The goal, then, is to reject the Ho and say Ha is correct. The Ho will always contain the condition of equality.
1

Formal Hypothesis Tests
In a formal hypothesis test, the opposite claims are given the names null hypothesis and alternative hypothesis . The
H
0
and the
H a
The null and alternative hypotheses need to be assigned as follows (see also p 367):
The
is the hypothesis being tested.
H
0
must:
 be the hypothesis we
 contain the condition of
( , , )
The
is always the
.
H a must:
 be the hypothesis we
 not contain the condition of equality ( , , )
A formal hypothesis test will always
to reject H
0
based on sample data, or the decision that there is not strong enough evidence to reject H
0
.
Example: A battery manufacturer claims that the average life of its batteries is at least
300 minutes. To test the claim, a sample of 100 batteries is drawn. The sample is tested and the mean battery life of the sample is 294 minutes with a sample standard deviation of 20 minutes. a) Write the null and alternative hypotheses. b) If the null hypothesis were rejected, what would you conclude? c) If the null hypothesis were not rejected, what would you conclude?
2

Example: A tire manufacturer claims its tires last on average more than 35,000 miles.
If we think the claim is false , then we would write the claim as H
0
, remembering to include the condition of equality. Our hypothesis for this test would be:
o
a
We would then hope that our sample data would allow the rejection of the null hypothesis, refuting the company’s claim.
Example: On the other hand, if we worked for the tire company and wanted to gather evidence to support their claim, then we would make the company’s claim H a
, and remember that equality should not be included in the claim. Our hypothesis test would then use the hypotheses:
o
a
If the sample data were able to support the rejection of H
0
, this would be strong evidence to support the claim H a
which is what the company believes to be true.
Note: For simplicity, in your text and on all quizzes and tests, it the condition of equality is inferred by the wording, that will automatically be the null hypothesis.
(Look at the example above and decide which would be the best “as worded”.)
3

Types of Error
Whenever sample data is used to make an estimate of a population parameter, there is always a probability of
. There are two main types of error that occur in hypothesis tests.
H
0
H
0
Summarizing:
H
0
True
H
0
False
H
0
Rejected
H
0
Not Rejected
Type I Error
Correct Decision
Correct Decision
Type II Error
Exercise: Go back and determine what a Type I Error and a Type II Error would be for our tire examples.
4

Level of Significance
In hypothesis tests, a
. That is, we want the probability of making a Type I Error to be small.
(We’d rather actually be right and think we’re wrong than think we’re right and actually be wrong.)
The maximum acceptable probability (of making a Type I error) is usually chosen at the beginning of the hypothesis test, and is called the
for the test. The level of significance is denoted by are
 
.10
,
 
.05
, and
 
.01

, and the most commonly used values
. Each of these represents the ________________
____________________________________________.
The probability of making a Type II Error in a hypothesis test is denoted by  . Once
 is determined, the value of  is also fixed, but the calculation of this value is beyond the scope of this course.
5

Types of Tests
There are
of hypothesis tests (see p 338 text, old book p 326):
The location of the “tail” is based on what we actually want to prove (H a
). The
HYPOTHESISED parameter (mean or proportion) will be “in the middle.”
– used when the null hypothesis being tested is a claim that the population parameter at least
( ) a given value. Note that the
– used when the null hypothesis being tested is a claim that the population parameter is at most
( ) a given value. Note that
– used when the null hypothesis being tested is a claim that the population parameter is equal to (=) a given value. Note that
( )
.
6

Given the three examples below, decide what type of test would be used: a LEFT-
TAILED, RIGHT-TAILED or TWO-TAILED test.
Example: The Census Bureau claims that the percentage of Tampa Area residents with a bachelor’s degree or higher is 24.4%. We would write the null and alternative hypotheses for this claim as:
H p

0
: .244
H a
: p

.244
In this case, we would have to reject H
0
if our sample percentage was either significantly more than 24.4%, or significantly less than 24.4%. That is, if our sample proportion was in ________________________ of the distribution of all sample proportions.
Example: From our first tire manufacturer example:
H
0
:
 
35, 000
H a
:
 
35, 000
We would reject
H
0
in the case above if our sample mean were significantly less than
35,000. That is, if our sample mean was in _____________________ of the distribution of all sample means.
Example: From our second tire manufacturer example:
H
0
:
 
35, 000
H a
:
 
35, 000
We would reject
H
0
in this case if our sample mean were significantly more than
35,000. That is, if our sample mean was in _______________________ of the distribution of all sample means.
7

7.2 Testing a Claim about the Mean Using a Large Sample
When a hypothesis test involves a claim about a population mean, then we will draw a sample and look at the sample mean to test the claim. If the sample drawn is large enough ( n

30
), then the Central Limit Theorem applies, and the distribution of sample means is approximately normal. As usual, we also have that:
 x
 
and
 x


 s n n
.
Note: Since s and n are known from the sample data, we have a good estimate of
 x
, but we do not know  since this is the parameter we are testing a claim about. In order to have a value for

, we will
in any hypothesis test.
Since the null hypothesis must be of one of the following types:
 
0
,
 
0
, or 
0 where

0
 
0 is a constant, we will always
8

The Standardized Test Statistic
There are three methods of determining if you should reject or not reject the null hypothesis:
Confidence intervals
(is the hypothesized statistic within the confidence interval for the sample)
P- values
(how probable it is that the sample statistic will be at/below/above the hypothesized parameter - that is, the probability of rejecting H
0
) This means that as the P-value gets smaller, the ________ likely we are to make a Type
I error (reject the null hypothesis when it is actually true.)
Critical Values
(how many standard deviations the sample is away from the hypothesized parameter) The more standard deviations from the hypothesized parameter, the ____________ likely we are to reject H
0
.
Note: The same conclusion should be made regardless of the method chosen.
Let’s go back to our previous example to see how this might work . . .
9

Example: A battery manufacturer claims that the average life of its batteries is at least
300 minutes (
 
300 ). To test the claim a sample of n = 100 batteries is drawn. The sample of batteries is tested and the mean battery life in the samples found to be x

294 minutes with a sample standard deviation of s = 20 minutes. Is this data sufficiently different from the manufacturer’s claim to justify rejecting the claim as false?
If the manufacturers claim is correct, then
 x

300 , and so we will assume that
 x

300 . From our sample, we can also estimate that
   x s n
20
100
The observed value x

294
has a z-score (standardized test statistic) of z
 x

 x
 x

2
 
3.0
Looking in the Standard Normal Table, we find that (

294)

.0013
.

2 .
Now, one of the following must be true:
Our assumption that
 
300 is incorrect.
OR
We have drawn a sample whose mean is so small that only 13 in 10,000 samples have a mean as low.
The likelihood of the second statement being true is quite small (.0013). Thus, we have strong evidence to believe that the first statement is true, and hence that the manufacturer overstated the mean lifetime of its batteries.
Now let’s reflect on this . . .
How could each of the following be represented in the example above?
P- values
Critical Values
10

Example: Suppose we believe that the mean body temperature of healthy adults is less than the commonly accepted measurement of 98.6
F. A sample of 60 healthy adults is drawn with an average temperature of x

98.2
F and with a sample standard deviation of s

1.1
.
Our hypotheses in this case would be:
H
0
H a
So we have a ________-tailed test with
 
0
98.6
.
Based on our sample data, our standardized test statistic is: z
 x
 
0 s / n

98.2 98.6


0.4
1.1/ 60 .142
 
2.82
How unusual is this? Do you think we should or should not reject the null hypothesis?
Exercise: Suppose we decide to make the stronger claim that the mean body temperature was less than 98.4. Find the new standardized test statistic.
Exercise: A researcher believes that the average commuting time for Tampa commuters is more than 25 minutes. A random sample of 100 Tampa commuters finds that the average commuting time is x

25.5
minutes with a standard deviation of s

11.5
minutes. If we wish to support the researcher’s claim
, determine the null and alternative hypotheses, the type of test, and the standardized test statistic.
11

The P-Value Method
The P-value (probability value) of a test is the probability of drawing a random sample whose standardized test statistic is at least as contrary to the claim of the null hypothesis as that observed in the sample group.
Example: In the hypothesis test for body temperature given above, we had:
H
0
:
 
98.6
H a
:
 
98.6
Our sample had a mean temperature of x

98.2
which is contrary to the null hypothesis.
Only a sample group with an average temperature less than 98.2 would be stronger evidence against H
0
. Thus the P-value of this test is (

98.2) .
Since the z-score of x is just our standardized test statistic z which has the Standard
Normal Distribution, (

98.2)

(
 
2.82)

.0024
.
Since the probability of drawing a sample as contrary to the null hypothesis as the observed sample (assuming
H
0
is true) is small, we would decide to reject
H
0
.
12

Calculating P-Values
In the example above, we calculated the P-value of the test by finding the area to the left of the standardized test statistic z on the standard normal curve. Notice that the example above was also a left-tailed test, and that any hypothesis test which is lefttailed will have the P-value calculated exactly as above.
(See diagram on p. 326 old 348 new)
Similarly, for a right-tailed test, we would calculate the
P-value by finding the area to the right of the standardized test statistic.
For a two-tailed test , the null hypothesis is always claiming that
0
, and so the sample data is contrary to this claim if the sample mean is either much higher or much lower than

0
. The P-value for a two-tailed test then is the area in both tails of the normal distribution more extreme than the standardized test statistic. Since the normal distribution is symmetric, this is just twice the area in one tail .
(See diagram on p. 354 old
386 new)
Example: The P-value of the battery life example
(p 10 notes)
would be:
(
 
3.0)

.0013
Example: A university claims that the average SAT score for its incoming freshmen is 1080. A sample of 56 freshmen at the university is drawn and the average SAT score is found to be x

1044 with a sample standard deviation of s

94.7
points.
Our hypotheses in this case would be:
H
0
H

So we have a _________-tailed test with
 
0
1080 .
Our standardized test statistic is: x
 z


0 s / n

94.7 / 56


36
12.65
 
2.85
Since the test is two-tailed, our P-value is given by:
 
2.85 or z

2.85) 2 (
 
2.85)
  
.0044
13

Exercise: Find the P-value for the test on Tampa commuting times
:
A researcher believes that the average commuting time for Tampa commuters is at least 25 minutes.
A random sample of 100 Tampa commuters finds that the average commuting time is x

25.5
minutes with a standard deviation of s

11.5
minutes. Find the P-value and think about what you might conclude. (See notes page 11.)
Deciding to Reject the Null Hypothesis
In the examples above, we saw that a very small P-value would lead us to reject the null hypothesis, and a high P-value would not.
Since the P-value of a test is the probability of randomly drawing a sample at least as contrary to
H
0
as the observed sample, we can also
H
0
The P-value then is the
Recall that the maximum acceptable probability of making a Type I Error is the level of significance  , and is usually determined at the outset of the hypothesis test.
The rule we will use to decide whether to reject H
0
is:
Reject H
0
if P
 
Do not reject H
0
if
P
 
Now go back to previous example and decide if we should reject the null hypothesis based on an alpha level of .01.
14

PROCESS for HYPOTHESIS TESTING:
a
o
o
a
o
Exercise: Is there sufficient evidence to support the researchers claim that the mean commuting time of Tampa commuters is more than 25 minutes at a  
.05
level of significance? Recall that the sample mean was 25.5 minutes, and the sample standard deviation was 11.5 min for a sample size of 100. Is there sufficient evidence to support a claim that the mean commuting time is more than 22 minutes at an  
.05
level of significance?
15

Rejection Regions and Critical Values
A second method to determine whether to reject the null hypothesis is to use rejection regions and critical values .
A rejection region for a hypothesis test is the range of values for the standardized test statistic that would cause us to decide to reject the null hypothesis.
0
The rejection region for a test is

.
For a left-tailed test, the rejection region is a region in the left tail of the normal distribution, for a right tailed test, it is in the right tail, and for a two tailed test, there are two equal rejection regions in either tail.
(See Diagram p. 341old 354 new)
Since 1- the level of confidence is the maximum acceptable probability of a Type I
Error (see t-distribution), we want the area under the normal curve in the rejection region to have an area of

. We can use this area to find our critical values.
(See Diagram p. 341 old, p 354 new)
16

Example: In our body temperature example, we were using a left-tailed test. If the level of significance was  
.05
, then the rejection region would be the values in the lowest 5% of the standard normal distribution. Looking up .05 in the standard normal table (or bottom of t-distribution chart), we see that this corresponds to a z-score of at most -1.645, so this would be our critical value z
0
, and so our rejection region is z
 
1.645
. Since our standardized test statistic z
 
2.82
falls into this rejection region, we would choose to reject
H
0
.
Example: In our SAT example, we had a two-tailed test, so the rejection region will be the two tails at either end of the normal distribution. If we again want  
.05
, then the area in both rejection regions together should be .05, and so we will look up
  in the standard normal table to get critical values of z
 
0
1.96
. The rejection region thus consists of: z
 
1.96
z
 
2.85
and z

1.96
. Since our standardized test statistic falls in the rejection region, we would reject the university’s claim that  
1080
.
Exercises: Find the critical values (z
0
)and rejection region(s):

A right-tailed test with
 
.05

A left-tailed test with  
.01

A two-tailed test with  
.10

A right-tailed test with
 
.02
17

Exercise: Mercury levels in fish are considered dangerous to people if they exceed 0.5 mg mercury per kilogram of meat. A sample of 50 tuna is collected, and the mean level of mercury in these 50 fish is 0.6 mg/kg with a standard deviation of 0.2 mg/kg.
A health warning will be issued if it the claim that the mean exceeds 0.5 mg/kg can be supported at the  
.10
level of significance.
Determine the null and alternative hypotheses in this case, the type of test, and the critical value(s) and rejection region.
Find the standardized test statistic for this data. Should the health warning be issued?
What is the P-value of this test?
Do p. 392 #
18

7.3 Testing a Claim about the Mean Using a Small Sample
Recall that if a sample of values for a normal random variable is drawn and if the sample size is less than 30, and the population standard deviation is unknown, then the random variable t
 x
  s / n has the Student t-distribution with n

1 degrees of freedom.
When testing a claim about the mean using sample data from a small sample, we should therefore use the appropriate t-distribution instead of the standard normal distribution to determine our standardized test statistic, critical values, rejection regions, and P-values.
The standardized test statistic for a t-distribution test will be given by: t
 x
 
0 s / n
Note: Probability values for the t-distributions are not given in the text, so we will limit our discussion to using the critical value method in this case. However, P-values for t-distribution tests can be calculated using the TI-83.
19

Finding Critical Values for the t-distributions
The process for locating the rejection regions for a t-distribution hypothesis test are the same as for the normal distribution tests. The critical values however will be different.
To find the critical value(s) t
0 for a test, first determine if the test is one-tailed or twotailed, and the level of significance  . The critical values can be found in the tdistribution table in the front of the book by looking up the entry in the column giving the level of significance and the row giving the degree of freedom.
Note that:

For a right-tailed test, t
0

For a left-tailed test, t
0 is the value in the table is the negative of the value in the table

For a two-tailed test, there are two critical values t
0
both the value and its opposite
Example: SAT Math Scores are normally distributed. A sample of scores for 16 students has mean score of x

522.8
with a sample standard deviation of s

154.5
.
Suppose we wished to support the claim that the average SAT Math score exceeds 500 using a  
.05
level of significance.
The null and alternative hypotheses in this case would be:
H
0
:
H a
:
So the test is ___________-tailed with 
0

500 .
Using the t-distribution table with One Tail
 
.05
and ______ degrees of freedom, we get a critical value of t
0
= . The rejection region is thus: t .
The standardized test statistic is given by: t
 x
 
0 s / n

522.8 500

22.8
154.5 / 16 38.625

.59
Because the standardized test statistic is _______ in the rejection region, we would
_______ reject H
0
, and so the sample data is _________ sufficient to support the claim that the mean exceeds 500 at the .05 level of significance.
20

Example: A biologist measures the weights of anesthetized female grizzly bears during winter. A sample of 14 bears is found to have a mean weight of x

376.6
lbs. with a sample standard deviation of s

32.5
lbs. Is there sufficient evidence to support the claim that the mean weight of all female bears in the area is less than 400 lbs. using an  
.01
level of significance? (Assume that the weights of female grizzly bears are normally distributed)
The null and alternative hypotheses in this case would be:
So the test is ________-tailed with 
0
 .
Using the t-distribution table with One Tail  
.01
get a critical value of t

0
and _____ degrees of freedom, we
. The rejection region is thus: t
The standardized test statistic is given by:

 
0

The standardized test statistic is/is not in the rejection region, so we would/would not reject H
0
, which supports/is insufficient to support the claim that the mean bear weight is less than 400 lbs.
Exercise: A statistics professor gives his class their second exam. In order to estimate the mean score, the professor chooses 8 exams at random and grades them. The mean grade for the eight exams chosen is x

84 and they have a standard deviation of
Assume that the students’ test scores are normally distributed. s

7 .
The professor believes the average score will be higher than the mean of 80 on the first exam. Conduct a hypothesis test using the
 
.05
professor’s claim.
level of significance, to support the
21

7.4 Testing a Claim about a Population Proportion
Recall that if a random sample is drawn and if the sample proportion ˆ p and ˆ 
5 , ˆ 5 , then the distribution of ˆ p is measured
is approximately normal with 
ˆ
 p and
  .
When testing a claim about a population proportion, the null hypothesis has one of the following forms: p
 p
0
So as with the mean, we will assume
, p
 p
0 p
 p
0
, or and p
 p
0 we will use the following standardized test statistic for a proportion test: z





ˆ  p
0 p q / n
Note: p
0 and q
0
are based on the hypothesis.
This random variable should have the standard normal distribution, and so we will calculate all of our rejection regions, critical values, and P-values using the standard normal distribution as we did when testing a mean using a large sample.
Example: A computer chip manufacturer tests microprocessors coming off the production line. In one sample of 577 processors, 37 were found to have defects. The company wants to claim that the proportion of processors that are defective is only 4%.
Can the company’s claim be rejected at the  
.01
level of significance?
The null and alternative hypotheses in this case would be:
H
0
: p

.04
H a
:
 
.04
So the test is two-tailed with p
0

.04
.
Since ˆ 37 / 577 .064
, the standardized test statistic is given by: z

ˆ  p
0 p q
0
/ n

(.04)(.96) / 577

.024
.008

3.0
Looking up z

3.00
in the standard normal table we get a value of .9987, so
(

3.00)
  
.0013
and since we have a two-tailed test, the P-value is just twice this amount or .0026. Since this is less than  
.01
, we can reject the company’s claim.
22

Example: An opinion poll of 1018 randomly chosen American adults finds that only
34% approve of the president’s job performance. (CBS News Poll Feb ’06.) The president’s political advisors want to know if this is sufficient data to show that less than 4/10 of Americans approve of the president’s job performance using the  
.05
level of significance.
The null and alternative hypotheses in this case would be:
H
0
H a

So the test is left-tailed with p

0
Since p
= 0.34, the standardized test statistic is given by: z =
For a left-tailed test with  
.05
, we will have z
0
 
1.645
, and so since z = z
0 the null hypothesis should not/be rejected, and so the data does/not support the claim p
< 0.4 at the  
.05
level of significance.
Exercise: Calculate the P-value of the above test.
Exercise: According to the IRS, 23.8% of all 2003 income tax returns were filed electronically. To estimate whether the number has increased, a random sample of 500 tax returns for 2004 is selected, and the IRS finds that 132 of the returns were filed electronically. Conduct a hypothesis test to support the claim that more than 23.8% of
2004 tax returns are filed electronically. What is the test statistic and P-value for the test at  
.01 and at =.05
? Write a conclusion and state why you do or do not reject H
0
.
23

Using the TI-83 to Conduct a Hypothesis Test: Mean – Large Sample
The program Z-Test on the TI-83 can be used to calculate the P-value and standardized test statistic for a hypothesis test of the mean ( n

30 ). To run the program:

Press [STAT]

Use the arrow keys to highlight the TESTS Menu

Highlight 1: Z-Test from the list and press [ENTER]
As before, if you know n ,

, and x , then make sure Inpt: is set to Stats. If you want the TI-83 to calculate from a data set, you would choose Data for Inpt:
Enter the values of

0
,

, x , and n .
Next choose the type of test by selecting the form of the alternative hypothesis
When you are finished highlight the word Calculate and press [ENTER]. The test statistic z , P-value, x , and n will then appear on the screen.
Example: Using the data from our temperature example:

0

98.6
 
1.1
x

98.2
, and n

60 , the Z-Test program gives:
P

.0024259384
z
 
2.816715161
with a P-value of
Using the TI-83 to Conduct a Hypothesis Test: Mean – Small Sample
The program T-Test on the TI-83 can be used to calculate the P-value and standardized test statistic for a tdistribution test. To run the program:

Press [STAT]

Use the arrow keys to highlight the TESTS Menu

Highlight 1: T-Test from the list and press [ENTER]
As before, if you know n , s , and x , then make sure Inpt: is set to Stats. If you want the TI-83 to calculate from a data set, you would choose Data for Inpt:
Enter the values of

0
, s , x , and n .
Choose the type of test as with the Z-Test program. And then Calculate.
The test statistic t , P-value, x , s , and n will then appear on the screen.
Example: Using the data from our SAT example:

0

500 s

154.5
x

522.8
, and n

16 , the T-Test program gives:
P

.2818916884
t

0.5902912621
with a P-value of
24

Using the TI-83 to Conduct a Hypothesis Test: Proportion
The program 1-PropZTest on the TI-83 can be used to calculate the P-value and standardized test statistic for a proportion test. To run the program:

Press [STAT]

Use the arrow keys to highlight the TESTS Menu

Highlight 5: 1-PropZTest from the list and press [ENTER]
Enter the values of p
0
, x , and n . As usual, if only ˆ is known, then x can be calculated as x
 ˆ
Next choose the type of test as with the other tests, and Calculate.
The test statistic z , P-value, ˆ , and n will then appear on the screen.
Example: Using the data from our processor example: p
0

.04
,
P
 x

37
.00310426
, and n

577 , the 1-PropZTest program gives: z

2.957234176
with a P-value of
25