Section 9.2 ~
Hypothesis Tests for Population Means
Introduction to Probability and Statistics
Ms. Young
Sec. 9.2
Objective

After this section you will understand and
interpret one- and two-tailed hypothesis
tests for claims made about population means,
and learn to recognize and avoid common
errors (type I and type II errors) in
hypothesis tests.
Sec. 9.2
Background Info

Recall that there are two possible outcomes of a
hypothesis test; to either reject or not reject the null
hypothesis

To determine whether to reject or not, a level of significance
(0.05 or 0.01) needs to be found


To find the level of significance, a P-value needs to be calculated
To calculate a P-value, you must first understand the
concepts of a normal distribution (introduced in ch.5):



Recall that if a distribution is normal, you can use z-scores
along with a z-score table to find probabilities of certain
values occurring
Also recall that a distribution begins to take the shape of a
normal distribution when the sample size is at least 30 and
becomes more and more normal as the sample size increases
(Central Limit Theorem)
In essence, a P-value (probability value) is the probability that
is found using z-scores and the z-score table

Be sure that you are using the sample standard deviation, s ,
when calculating the z-score since you are comparing a sample
(group mean or group proportion) to the entire population
Sec. 9.2
One-Tailed Hypothesis Tests

As mentioned in section 9.1, hypothesis tests can either be one-tailed
(left or right) or two-tailed


The process for conducting a left-tailed test is the same as the process for
conducting a right-tailed test, but a two-tailed test varies slightly
Example 1 ~ Left-Tailed Hypothesis Test:

Columbia College advertises that the mean starting salary of its
graduates is $39,000. The Committee for Truth in Advertising
suspects that this claim is exaggerated and that the mean starting
salary for graduates is actually lower. They decide to conduct a
hypothesis test to seek evidence to support this suspicion.
 Step 1: State the null and alternative hypotheses
H a :   $39,000
H 0 :   $39,000
 Step 2: Draw a sample and come up with a sample statistic and
the standard deviation of that sample:


Suppose that the committee gathered a sample of 100 graduates and the
sample mean is x  $37, 000
Also suppose that the standard deviation for that sample is $6,150
Sec. 9.2

Example 1 Cont’d…

Step 3: Calculate the P-value (using the normal distribution and
z-scores) and determine the level of significance

z
In order to calculate the P-value, we need to find the z-score using the Central
Limit Theorem since we are dealing with the mean of a group. Since we do not
know the population standard deviation, we will use the standard deviation
found for the sample as an estimate.
x 
 n

z
37, 000  39, 000
6150
100

z  3.25
 Using the z-score table we find that a z-score of -3.25 correlates with a
probability of .0006, or .06%. This is the P-value.
 Since this value is less than .05 it is significant at the .05 level, but even
better, this value is less than .01 which means that it is significant at the
.01 level

Step 4: Decide if you should reject or not reject the null
hypothesis


Since the P-value is significant at both levels (.05 and .01), we should reject
the null hypothesis of $39,000
What this means is that we have strong evidence to believe that Columbia
College exaggerated about the mean starting salary of their graduates being
$39,000 and that it is most likely lower.
Sec. 9.2
One-Tailed Hypothesis Tests
Example 2 ~ Right-Tailed Hypothesis Test

In the United States, the average car is driven about 12,000 miles each year. The owner
of a large rental car company suspects that for his fleet, the mean distance is greater
than 12,000 miles each year. He selects a random sample of n = 225 cars from his fleet
and finds that the mean annual mileage for this sample is x  12,375 miles. Suppose that
the standard deviation for that sample is 2,415 miles. Interpret this claim by conducting
a hypothesis test.

Step 1: State the null and alternative hypotheses
H 0 :   12, 000 miles


H a :   12,000
Step 2: Draw a sample and come up with a sample statistic and the standard
deviation of that sample
 This information was already given:
 The sample mean is x  12,375
 The standard deviation for that sample is 2,415 miles
Step 3: Calculate the P-value and determine the level of significance:
 The z-score is:
z
12,375  12, 000
 2.33
2415
225
Sec. 9.2
One-Tailed Hypothesis Tests
Example 2 Cont’d…

Step 3 cont’d…



The z-score was found to be 2.33 which corresponds to a probability of .9901
on the z-score table, but that represents the area below 12,375 and we are
interested in knowing the probability of a car being driven more than that value
so we subtract .9901 from 1 (1 - .9901) and get a probability of .0099
The P-value is .0099 which is less than .01, meaning that it is significant at the
.01 level
Step 4: Decide if you should reject or not reject the null hypothesis


Since the P-value is significant at both levels (.05 and .01), we should reject
the null hypothesis of 12,000 miles
What this means is that we have strong evidence to believe that the mean
distance traveled for the rental car fleet is greater than 12,000 miles
Sec. 9.2
Critical Values for Statistical Significance


Since we can decide to reject the null hypothesis if the P-value is .05 or
lower (or .01 or lower), we can use critical values as a quick guideline to
decide if we should reject the null hypothesis or not
Critical values for .05 significance level:
For a left-tailed test, the z-score that corresponds to a probability of .05 is
-1.645, so any z-score that is less than or equal to -1.645 will be statistically
significant at the .05 level
 For a right-tailed test, the z-score that corresponds to a probability of .05
(which we would look for .95 on the chart) is 1.645, so any z-score greater
than or equal to 1.645 will be statistically significant at the .05 level


Critical values for the .01 significance level:
For a left-tailed test, the z-score that corresponds to a probability of .01 is
-2.33, so any z-score that is less than or equal to -2.33 will be statistically
significant at the .01 level
 For a right-tailed test, the z-score that corresponds to a probability of .01
(which we would look for .99 on the chart) is 2.33, so any z-score greater than
or equal to 2.33 will be statistically significant at the .01 level

Sec. 9.2
Critical Values for Statistical Significance
Sec. 9.2
Two-Tailed Hypothesis Tests


The process for conducting a two-tailed hypothesis test is very similar to
the one-tailed tests, except the critical values are slightly different
Since a two tailed test tests both above and below the claimed value, a
.05 significance level would have to be split between the two extremes
thus looking for a z-score that corresponds to a probability of .025

The z-scores that correspond to a probability of .025 are -1.96 and
1.96, so for a two-tailed test, it is significant at the .05 level if the
z-score is less than or equal to -1.96 or greater than or equal to 1.96
Sec. 9.2
Two-Tailed Hypothesis Tests

For a two-tailed test, a .01 significance level would mean that the
z-score needs to correspond to a probability of .005 (.01 split in
half)


The z-scores that correspond to a probability of .005 are -2.575 and
2.575, so if the z-score is less than or equal to -2.575 or greater
than or equal to 2.575, then it is statistically significant at the .01
level
Summary of critical values for two-tailed tests:

.05 significance level:
z  1.96 or z  1.96

.01 significance level:
z  2.575 or z  2.575
Sec. 9.2
Two-Tailed Hypothesis Tests
Example 3 ~ Two-Tailed Hypothesis Test:

Consider the study in which University of Maryland researchers
measured body temperatures in a sample of n = 106 healthy adults,
finding a sample mean body temperature of x  98.20°F with a
sample standard deviation of 0.62°F. We will assume that the
population standard deviation is the same as the standard deviation
found from the sample. Determine whether this sample provides
evidence for rejecting the common belief that the mean human body
temperature is   98.60°F

Step 1: State the null and alternative hypotheses
H 0 :   98.6°F

H a :   98.6°F
Step 2: Draw a sample and come up with a sample statistic and the
standard deviation of that sample
 This information was already given:
 The sample mean is x  98.20°F
 The standard deviation for that sample is 0.62°F
Sec. 9.2
Two-Tailed Hypothesis Tests
Example 3 Cont’d…

Step 3: Calculate the P-value and determine the level of significance


To calculate the P-value for a two tailed test, you must find the z-score
like you would with a one-tailed test, but the probability that corresponds
to it must then be multiplied by 2
The z-score is:
z


98.2  98.6
 6.64
0.62
106
The P-value is less than .02 (.01 * 2), and since the z-score of -6.64 is
significantly lower than -1.96 and -2.575, this would be statistically significant
at both levels
Step 4: Decide if you should reject or not reject the null hypothesis

The null hypothesis should be rejected which provides strong evidence that
the mean human body temperature is not 98.6°F. It may be either higher or
lower.
Sec. 9.2
Common Errors in Hypothesis Testing

A hypothesis test guides us in the right direction of either
supporting or not supporting a claim, but even if a hypothesis test
is carried out correctly, two common types of errors may affect
the conclusions

Type I error ~ when the null hypothesis is wrongly rejected

Ex. ~ Consider a drug company that seeks to be sure that its “500milligram” aspirin tablets really contain 500 milligrams. If the tablets
contain less than 500 milligrams, the consumers are not getting the
advertised dose. If the tablets contain more than 500 milligrams, the
consumers are getting more than the advertised dose (which could be
potentially dangerous).


Suppose that a hypothesis test gave evidence to reject the null hypothesis
(thus stating that the drug does not contain 500 milligrams, when in reality it
does contain 500 milligrams
 This would be an example of a type I error; it is appearing as though the
drug has either more or less aspirin than it should, when in reality it has
the correct amount
Type II error ~ when the null hypothesis should be rejected, but it
wasn’t

Consider the drug scenario above.

Suppose that a hypothesis test gave evidence to not reject the null hypothesis,
stating that the drug does contain 500 milligrams, when in reality it does not
contain 500 milligrams
 This would be an example of a type II error; it is appearing as though
the drug has the correct amount of aspirin when in reality it doesn’t
Sec. 9.2
Common Errors in Hypothesis Testing Cont’d…

Example 4 ~ Type I and Type II Errors

The success of precious metal mines depends on the purity (or grade)
of ore removed and the market price for the metal. Suppose the
purity of gold ore must be at least 0.5 ounce of gold per ton of ore in
order to keep the mine open. Samples of gold ore are used to
estimate the purity of the ore for the entire mine. Discuss the
impact of type I and type II errors on two of the possible
alternative hypotheses:
H a : purity < 0.5 ounce per ton
H a : purity > 0.5 ounce per ton

For the first alternative hypothesis H a : purity < 0.5 ounce per ton


Wrongly rejecting the null hypothesis (type I error) would result in closing the
mine, when in reality they were up to standard
 This would be a terrible loss for the mine and all workers
Wrongly “accepting” the null hypothesis (type II error) would result in the
mine staying open when they should really be closed
 This would impact the consumers in that they are not getting the quality
in which they believe they are getting
Sec. 9.2
Common Errors in Hypothesis Testing Cont’d…

Example 4 ~ Type I and Type II Errors

The success of precious metal mines depends on the purity (or grade)
of ore removed and the market price for the metal. Suppose the
purity of gold ore must be at least 0.5 ounce of gold per ton of ore in
order to keep the mine open. Samples of gold ore are used to
estimate the purity of the ore for the entire mine. Discuss the
impact of type I and type II errors on two of the possible
alternative hypotheses:
H a : purity < 0.5 ounce per ton
H a : purity > 0.5 ounce per ton

For the second alternative hypothesis H a : purity > 0.5 ounce per ton


Wrongly rejecting the null hypothesis (type I error) would result in the mine
staying open when it shouldn’t
Wrongly accepting the null hypothesis (type II error) would result in the mine
closing when it should be able to stay open