Hypothesis Testing of Single Mean
This module provides examples of Hypothesis Testing of a Single Mean and a Single Proportion as apart of the Collaborative Statistics collection (col10522) by Barbara Illowsky and Susan Dean.
Jeffrey, as an eight-year old, established a mean 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 ofexpensive goggles and timed Jeffrey for 15 25-yard freestyle swims. For the 15 swims, Jeffrey’smean 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 preset . Assume thatthe swim times for the 25-yard freestyle are normal.
Set up the Hypothesis Test:
Since the problem is about a mean, this is a test of a single population mean.
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 that there is no evidence to conclude that Jeffrey swims the 25-yard
free-style, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yardfree-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the nullhypothesis is false.)
A college football coach thought that his players could bench press a mean weight of 275
pounds. It is known that the standard deviation is 55 pounds. Three of his players thought
that the mean weight 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: datafrom Reuben Davis, Kraig Evans, and Scott Gunderson.)
Conduct a hypothesis test using a 2.5% level of significance to determine if the bench press
mean is more than 275 pounds.
Statistics students believe that the mean score on the first statistics test is 65. A statistics
instructor thinks the mean score is higher than 65. He samples ten statistics students and obtainsthe scores 65; 65; 70; 67; 66; 63; 63; 68; 72; 71. He performs a hypothesis test using a 5% level ofsignificance. The data are from a normal distribution.
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.
note: Hypothesis testing problems consist of multiple steps. To help you do the problems, solutionsheets are provided for your use. Look in the Table of Contents Appendix for the topic “SolutionSheets.” If you like, use copies of the appropriate solution sheet for homework problems.
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!
Before his shampoo
I counted 42.
At the end of his bath,
I redid the math
And the new shampoo had killed 17 fleas.
So now 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?
Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence
that the percentage of fleas 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 ConfidenceInterval.
Confidence Interval: (0:26; 0:55) We are 95% confident that the true population proportion
pof fleas that are killed by the new shampoo is between 26% and 55%.
note: This test result is not very definitive since the p-value is very close to alpha. In reality, one
would probably do more tests by giving the dog another bath after the fleas have had a chance to
Test of μ = 16.43 vs< 16.43
The assumed standard deviation = 0.8
N Mean SE Mean 95% Upper Bound Z P
15 16.000 0.207 16.340 -2.08 0.019
conclusion: Rejected at alpha=0.05
.One-Sample Z: C1
Test of μ = 275 vs> 275
The assumed standard deviation = 55
Variable N Mean StDev SE Mean 95% Lower Bound Z P
C1 30 286.2 55.9 10.0 269.6 1.11 0.133
Conclusion: Accepted at alpha=.025
One-Sample T: C2
Test of μ = 65 vs> 65
Variable N Mean StDev SE Mean 95% Lower Bound T P
C2 10 67.00 3.20 1.01 65.15 1.98 0.040
4.Test of p = 0.5 vs p ≠ 0.5
Sample X N Sample p 95% CI P-Value
1 53 100 0.530000 (0.427581, 0.630595) 0.617
5.Test of p = 0.3 vs p ≠ 0.3
Sample X N Sample p 95% CI P-Value
1 43 150 0.286667 (0.215861, 0.366109) 0.724