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Hypothesis Testing of Single Mean and Single Proportion: Examples
Summary: This module provides a examples of Hypothesis Testing of Single Mean and Single Proportion as a part of Collaborative Statistics collection (col10522) by Barbara Illowsky and Susan Dean.
Example 1
Problem 1
Jeffrey, as an eight-year old, established an average time of 16.43
seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds.
His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster by using
goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15
25-yard freestyle swims. For the 15 swims, Jeffrey's average time was 16 seconds.
Frank thought that the goggles helped Jeffrey to swim faster than the 16.43
seconds. Conduct a hypothesis test using a preconceived
Solution 1
Setting up the Hypothesis Test:
Since the problem is about a mean (average), this is a test of a single population mean.
For Jeffrey to swim faster, his time will be
less than 16.43 seconds. The "
Calculating the distribution needed:
Random variable:
Distribution for the test:
Calculate the p-value using the normal distribution for a mean:
Graph:
 |
Interpretation of the p-value: If
Compare
Make a decision: Since
This means that you reject
Conclusion: At the 5% significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds.
The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:
Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Stats
and press ENTER. Arrow down and enter 16.43 for ENTER. Arrow down to Calculate and press
ENTER. The calculator not only calculates the p-value (Draw
(instead of Calculate). Press ENTER. A shaded graph appears with Draw that no
other equations are highlighted in
When the calculator does a Z-Test, the Z-Test function finds the p-value by
doing a normal probability calculation using the Central Limit Theorem
(Chapter 7):
2nd DISTR normcdf
The Type I and Type II errors for this problem are as follows:
The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)
The Type II error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)
Historical Note: The traditional way to compare the two probabilities,
 |
Example 2
Problem 1
A college football coach thought that his players could bench press an average of 275 pounds. It is known that the standard deviation is 55 pounds. Three of his players thought that the average was more than that amount. They asked 30 of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual different weights were (frequencies are in parentheses) 205(3); 215(3); 225(1); 241(2); 252(2); 265(2); 275(2); 313(2); 316(5); 338(2); 341(1); 345(2); 368(2); 385(1). (Source: data from Reuben Davis, Kraig Evans, and Scott Gunderson.)
Conduct a hypothesis test using a 2.5% level of significance to determine if the bench press average is more than 275 pounds.
Solution 1
Setting up the Hypothesis Test:
Since the problem is about a mean (average), this is a test of a single population mean.
Calculating the distribution needed:
Random variable:
Distribution for the test: It is normal because
Calculate the p-value using the normal distribution for a mean:
Interpretation of the p-value: If
 |
Compare
Make a decision: Since
Conclusion: At the 2.5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds.
The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:
Put the data and frequencies into lists. Press
STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Data and
press ENTER. Arrow down and enter 275 for ENTER.
Arrow down to Calculate and press ENTER. The calculator not only
calculates the p-value (Draw (instead of Calculate). Press ENTER. A shaded graph appears with Draw that no other equations are highlighted in
Example 3
Problem 1
Statistics students believe that the average score on the first statistics test is 65. A statistics instructor thinks the average score is higher than 65. He samples ten statistics students and obtains the scores 65; 65; 70; 67; 66; 63; 63; 68; 72; 71. He performs a hypothesis test using a 5% level of significance. The data are from a normal distribution.
Solution 1
Setting up the Hypothesis Test:
A 5% level of significance means that
Since the instructor thinks the average score is
higher, use a "
Calculating the distribution needed:
Random variable:
Distribution for the test: If you read the problem carefully, you will notice that there is
no population standard deviation given. You are only given
Use
Calculate the p-value using the Student-t distribution:
Interpretation of the p-value: If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 67 pounds or more.
 |
Compare
Since
Make a decision: Since
This means you reject
Conclusion: At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.
The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:
Put the data into a list. Press STAT and arrow over to
TESTS. Press 2:T-Test. Arrow over to Data and press ENTER. Arrow down
and enter 65 for Freq:.
Arrow down to ENTER. Arrow down to Calculate and press ENTER. The calculator not
only calculates the p-value (Draw (instead of Calculate). Press ENTER. A shaded graph
appears with Draw that no other equations are highlighted in
Example 4
Problem 1
Joon believes that 50% of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is the same or different from 50%. Joon samples 100 first-time brides and 53 reply that they are younger than their grooms. For the hypothesis test, she uses a 1% level of significance.
Solution 1
Setting up the Hypothesis Test:
The 1% level of significance means that
The words "is the same or different from" tell you this is a two-tailed test.
Calculating the distribution needed:
Random variable:
Distribution for the test: The problem contains no mention of an average. The
information is given in terms of percentages. Use the distribution for
Calculate the p-value using the normal distribution for proportions:
where
Interpretation of the p-value: If the null hypothesis is true, there is 0.5485 probability
(54.85%) that the sample (estimated) proportion
 |
Compare
Since
Make a decision: Since
Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides who are younger than their grooms is different from 50%.
The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:
Press STAT and arrow over to TESTS. Press 5:1-PropZTest.
Enter .5 for Prop and arrow to not equals ENTER. Arrow down to Calculate and press ENTER. The calculator
calculates the p-value (Prop not
equals .5 is the alternate hypothesis. Do this set of instructions again except
arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears
with Draw that no other equations are highlighted in
The Type I and Type II errors are as follows:
The Type I error is to conclude that the proportion of first-time brides that are younger than their grooms is 50% when, in fact, the proportion is actually different from 50%. (Reject the null hypothesis when the null htpothesis is true).
The Type II error is to conclude that the proportion of first-time brides that are younger than their grooms is different from 50% when, in fact, the proportion is actually 50%. (Do not reject the null hypothesis when the null hypothesis is false.)
Example 5
Problem 1
Suppose the proportion of households that have three telephone numbers is 30%. The telephone company has reason to believe that the proportion of households is less than 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three telephone numbers.
Solution 1
Setting up the Hypothesis Test:
Calculating the distribution needed:
The random variable is
The distribution for the hypothesis test is
Exercise 1
Calculate the p-value using the normal distribution for proportions:
Solution 1
p' =
Calculate the p-value using the normal distribution for proportions:
The value that helps determine the p-value is
Exercise 2
What is a success for this problem?
Solution 2
A success is having three telephone numbers in your household
Exercise 3
Solution 3
Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately.
Compare
Fill in the blanks.
Exercise 4
p-value = _________
Solution 4
p-value = 0.3594
Exercise 5
Make a decision.
_____________(Reject/Do not reject)
Solution 5
Assuming that
Write a conclusion.
The next example is a poem written by a statistics student named Nicole Hart. The
solution to the problem follows the poem. Notice that the hypothesis test is for a single
population proportion. This means that the null and alternate hypotheses use the
parameter
Note:
Notice the solution sheet that has the solution. Look in the Table of Contents for the topic "Solution Sheets." Use copies of the appropriate solution sheet for homework problems.Example 6
Problem 1
My dog has so many fleas,
They do not come off with ease.
As for shampoo, I have tried many types
Even one called Bubble Hype,
Which only killed 25% of the fleas,
Unfortunately I was not pleased.
I've used all kinds of soap,
Until I had give up hope
Until one day I saw
An ad that put me in awe.
A shampoo used for dogs
Called GOOD ENOUGH to Clean a Hog
Guaranteed to kill more fleas.
I gave Fido a bath
And after doing the math
His number of fleas
Started dropping by 3's!
With his old shampoo
I counted 42.
At the end of his bath,
I redid the math
And the new shampoo had killed 17 fleas.
So no I was pleased.
Now it is time for you to have some fun
With the level of significance being .01,
You must help me figure out
Use the new shampoo or go without?Solution 1
Setting up the Hypothesis Test:
Calculating the distribution needed:
In words, CLEARLY state what your random variable
State the distribution to use for the test.
Normal:
Test Statistic:
Calculate the p-value using the normal distribution for proportions:
In 1 – 2 complete sentences, explain what the p-value means for this problem.
If the null hypothesis is true (the proportion is 0.25), then there is a 0.0103 probability that the sample (estimated)
proportion is 0.4048
Use the previous information to sketch a picture of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the p-value.
 |
Compare
Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using COMPLETE SENTENCES.
| alpha | decision | reason for decision |
|---|---|---|
| 0.01 | Do not reject |
Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of flea that are killed by the new shampoo is more than 25%.
Construct a 95% Confidence Interval for the true mean or proportion. Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the Confidence Interval.
 |
Confidence Interval:
Note:
This is a weak test since the p-value is very close to alpha. In reality, one would probably do more tests.Glossary
- Central Limit Theorem:
- Given a random variable (RV) with known mean
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