COC Math 140 Chapter 9 Hypothesis Testing for Means (one and two) Practice Problems
Perform an hypothesis test for each of the following. Be sure to (1) state your null and alternative hypotheses & define your parameters; (2) write your conclusion, which must include your decision (in terms of the null hypothesis), alpha level, p-value, and context (in terms of the alternative hypothesis). You may assume that conditions have been met and checked.
1. A battery company wants to test Ho: µ = 30 versus Ha: µ > 30 based on an SRS of 15 new batteries with mean lifetime x = 33.9 hours and standard deviation sx = 9.8 hours. Run the hypothesis test and determine if we can conclude that the battery life of all batteries like these is more than 30 hours.
Ho: m = 30 Ha: m > 30
where m is the true, unknown mean lifetime in hours for all new batteries like these.
Fail to reject. With a p-value of over 7% and an alpha of 5%, we do not have sufficient evidence to show that the battery life for all new batteries like these is greater than 30 hours.
2. The level of dissolved oxygen (DO) in a stream or river is an important indicator of the water’s ability to support aquatic life. A researcher measures the DO level at 15 randomly chosen locations along a stream. Here are the results in milligrams per liter (mg/l):
4.53 / 5.04 / 3.29 / 5.23 / 4.13 / 5.50 / 4.83 / 4.405.42 / 6.38 / 4.01 / 4.66 / 2.87 / 5.73 / 5.55
A dissolved oxygen level below 5 mg/l puts aquatic life at risk.
Do we have convincing evidence at the alpha level of 0.05 significance level that aquatic life in this stream is at risk?
Ho: m = 5 Ha: m < 5
Where m is the true, unknown mean DO level in mg/l in the entire stream.
Fail to reject Ho. With an alpha level of 5% and a p-value of over 18%, we do not have sufficient evidence to show that DO levels are below 5 mg/l in this stream.
3. Every road has one at some point – construction zones that have much lower speed limits. To see if drivers obey these lower speed limits, a police officer uses a radar gun to measure the speed (in miles per hour, or mph) of a random sample of 10 drivers in a 25 mph construction zone. Here are the data:
27 / 33 / 32 / 21 / 30 / 30 / 29 / 25 / 27 / 34Is there convincing evidence that the average speed of drivers in this construction zone is greater than the posted speed limit?
Ho: m = 25 Ha: m > 25
Where m is the true, unknown mean speed (in mph) for all cars driving through this construction zone.
Reject Ho. At an alpha level of 5% and a p-value of less than 1%, we have sufficient evidence to conclude that the mean speed in mph for all cars driving through this construction zone is greater than 25 mph.
4. Poisoning by the pesticide DDT causes convulsions in humans and other mammals. Researchers seek to understand how the convulsions are caused. In a randomized comparative experiment, they compared 6 white rats poisoned with DDT with a control group of 6 un-poisoned rats. Electrical measurements of nerve activity are the main clue to the nature of DDT poisoning. When a nerve is stimulated, its electrical response shows a sharp spike followed by a much smaller second spike. The researchers measured the height of the second spike as a percent of the first spike when a nerve in a rat’s leg was stimulated. For the poisoned rats, the results were
12.207 / 16.869 / 25.050 / 22.429 / 8.456 / 20.579The control group data were:
11.074 / 9.686 / 12.064 / 9.351 / 8.182 / 6.642Do these data provide convincing evidence that DDT affects the mean relative height of the second spike’s electrical response?
Ho: m(poisoned) = m(control) Ha: m(poisoned) ¹ m(control)
Where m(poisoned) is the true mean relative height of the second spike’s electrical response for all rats like these and m(control) is the true mean relative height of the second spike’s electrical response for all rats like these.
Reject Ho. With a p-value of about 2% and an alpha level of 5%, we have sufficient evidence to conclude that the true mean relative height of the second spike’s electrical response for all rats like these that are poisoned are not the same as the control rats.
5. Researchers equipped random samples of 56 male and 56 female students from a large university with a small device that secretly records sound for a random 30 seconds during each 12.5-minute period over two days. Then they counted the number of words spoken by each subject during each recording period and, from this, estimated how many words per day each subject speaks. The female estimates had a mean of 16,177 words per day with a standard deviation of 7520 words per day. For the male estimates, the mean was 16,569 and the standard deviation was 9108. Do these data provide convincing evidence of a difference in the average number of words spoken in a day by male and female students at this university?
Ho: m(males) = m(females) Ha: m(males) ¹m(females)
Where m(males) is the true population mean for total number of words spoken for all males like these per day; and m (females) is the true population mean for total number of words spoken for all females like these per day.
Fail to reject. With a p-value more than any reasonable alpha level, we do not have sufficient evidence to conclude that all males and females like these speak a different mean number of words daily.
6. A company that makes hotel toilets claims that its new pressure-assisted toilet reduces the average amount of water used when compared to its current model. To test this claim, the company randomly selects 30 toilets of each type and measures the amount of water that is used when each toilet is flushed once. For the current model toilets, the mean amount of water used is 1.64 gallons with a standard deviation of 0.29 gallons. For the new toilets, the mean amount of water used is 1.59 gallons with a standard deviation of 0.18 gallons. Carry out an appropriate significance test. What conclusion would you draw?
Ho: m(new) = m(current) Ha: m(new) < m(current)
Where m(new) is the true population parameter mean amount of water used for all toilets like these and m(current) is the true population parameter mean amount of water used for all toilets like these.
Fail to reject Ho. With a p-value well over any reasonable alpha, we do not have sufficient evidence to support that the new toilets reduce the average amount of water used when compared to the current model.