Practice Exam Populations,Samples, and Confidence Intervals
Practice Problems using Hypothesis tests
1. A nylon rope maker claims its repelling rope's average breaking strength is at least 800 pounds. A researcher sets up a hypothesis test regarding the breaking strength of the rope where Ho: μ ≤ 800. where α = .05.
- Rewrite his null and alternative hypotheses as complete sentences. What does the hypothesis suggest about where the benefit of the doubt is being placed? Explain your answer.
The null states that the average breaking strength is less than (no more than) 800 pounds. This assumes that the rope is weak, and puts the burden of proof on the rope.
The alternative is that the rope will hold more than 800 pounds. (the rope is strong)
b. What does it mean to make a type I error, and, specifically, what would constitute a type one error in this case?
A type I error is believing the alternative when it is false. In this case we believe the rope is strong when it is not.
- In this case, is it appropriate to use α = .05? Explain why or why not.
so we have a 5% chance that we believe the rope is strong when it isn’t. This should be lower, if making this mistake could result in personal injury.
- what would we conclude if the sample average was much larger than 800 pounds?
We would believe the rope is strong
- what would we conclude if the sample average was slightly less than 800 pounds?
We would believe the rope was weak
- what if the sample average is much less than 800?
We would believe the rope was weak
1. Waders Inc. currently has an exclusive contract with United Airlines for all domestic air travel. In return for the exclusive contract, United gives the company a 20 percent discount. After the discount, an average round-trip flight on United costs the company $687 with a standard deviation of $120. Sam Innair is the United representative that handles the Waders Inc. Account.
Sally Flibinite represents a discount airline that is trying to get the Waders Inc. account. "Give us a try, she says, I guarantee we'll save you money." Waders Inc agrees to a test. In a sample of 49 flights, the discount airline's average fare was $650 with a standard deviation of $140.
Upon seeing the results, Sally proclaims the test a success.
When Waders Inc. presents United with the test results, Sam claims that you might find the same average over a sample of their own flights, and so the results are inconclusive.
Ho : μ ≥ 687 (Sally’s airline is at least as costly as United)
H1 : μ < 687 (Sally’s airline is cheaper)
Since we are trying to show that sally is cheaper, that has to be the alternative.
.10Z = (650 – 687)/120√49= -2.16
Reject Ho
-2.16 -1.28 687
We are using a z because the standard deviation came from a population. (Testing Sam’s claim that united could have a sample of 650 based on their average and standard deviation)*******
We can be 90 percent sure that sally is cheaper. (We could not be 99% sure of this)
.01538Z = (650 – 687)/120√49= -2.16
Reject Ho
-2.16 687
so we can be (1-.01538) = 98.46 percent sure that Sally is cheaper
**** This problem could have been solved with a t if you used s=140 instead of σ =120.
2. Last year Alert America’s residential alarm system sales reps, sold an average of 50 systems per salesperson. This year sales seem slower. As with most sales forces of this type, employee turnover has been brisk, but this year the problem seems worse than usual. Last year, the average tenure (time with the company) of an Alert America sales representative was 2.9 years with a standard deviation of 2 years. When the company inventoried tenure in May, the average of a random sample of thirty-six sales representatives was only 2 years, but the standard deviation was 3 years. Management opinion is divided as to what to make of the difference. Some think that this is just a temporary abnormality, while others think that this is evidence of an underlying problem. As chief of statistical analysis, each side turns to you for your opinion.
Ho : μ ≥ 2.9 (Tenure has not decreased)
H1 : μ < 2.9 (Tenure has decreased)
Since we are trying to show that something has changed, that has to be the alternative.
.10Z = (2 – 2.9)/2√36= -2.70
Reject Ho
-2.7 -1.28 2.9
We are using a z because the standard deviation came from a population.*******
We can be 90 percent sure that tenure has decreased.
The p-value is:
since the p-value is .3% we can be 99.7 percent sure that tenure has decreased
**** This problem could have been solved with a t if you used s=3 instead of σ =2.
3. McDonalds has been considering dropping the pickles from its burgers. There is some sentiment that the ubiquitous pickle adds to the cost of the sandwich, but most people remove them anyway. The CEO decides that if we can be reasonably certain more than two-thirds of customers remove the pickle, then the pickle should no longer be placed on the sandwich (except by request). Ronald was enlisted to spy on a random sample of one thousand customers. He found that 69 percent removed the pickle. What should the company do and why?
Proportions:
Ho : π ≤ .67 (fewer than 2/3 remove the pickle: Keep the pickle)
H1 : π > .67 (more than 2/3 remove the pickle: get rid of the pickle)
Since we are trying to show that people don’t like the pickle, that has to be the alternative.
Z = (.69 - .67)/√(.33)(.67)/1000)=1.345
.01Reject Ho
.67 1.345 2.33
We can NOT be at least 99 percent sure that more than 2/3 remove the pickle. (though we could be 90 percent sure). The p-value is 1-.91 = .089 so we can be 91 percent sure that more than 2/3 remove the pickle.
In this case we need definitive evidence. After all the pickle is ubiquitous.
Mc Donalds is also considering changing pickle suppliers. With the current supplier, each McD’s pays an average of 1 cent per pickle. A rival in a sample of 49 pickle prices from a rival supplier (pickle prices vary by region due to shipping costs etc.) The average price was .9 cents per pickle with a standard deviation of .02 cents. Should we switch suppliers? Explain why or why not.
Ho : μ ≥ 1 (New vendor is more expensive)
H1 : μ < 1 (New vendor is less expensive)
Since we are trying to show that the new vendor is better, that has to be the alternative.
.10TS = (.9 – 1)/.02√49= -35
Reject Ho
-35 -2.423 10.5
We are using a t because the standard deviation came from a sample.
We can be 90 percent sure that the new vendor is better. The new vendor ‘s sample is 35 standard errors below 1 cent. This is a huge difference, but that is because the standard deviation (.02 cents) is so small. (If I were working this problem, I would investigate that number. It is possible that it should have read 2 cents.
The p-vlaue is: .00000000000000000000000000000000000000039!
- Pennsylvania Governor Tom Ridge has a longstanding reputation for being "tough on crime." His opponent is currently running ads that dispute this. Before Ridge took office, the average criminal spent 10.5 years in prison before parole. Ridge's opponent says that a random sample of forty-nine recent parolees spent an average of only ten years in prison (the standard deviation in the sample was fourteen years). Assuming you work for an independent newspaper, evaluate the claim made by the Ridge opponent. What would you report to your readers?
Ho : μ ≥ 10.5 (Ridge is no worse than predecessors)
H1 : μ < 10.5 (Ridge is worse than predecessors)
Since we are trying to show that Ridge is worse, that has to be the alternative.
.10TS = (10 – 10.5)/14/√49= -.25
Reject Ho
-1.303 -.25 10.5
We are using a t because the standard deviation came from a sample.
We can NOT be 90 percent sure that Ridge is worse. Ridge is only .25 standard errors below the average. Substantial evidence of being soft on crime would be finding a sample average that was more than 2.423 standard errors below average. The p-value is .40, so we can only be 60 percent sure that Ridge is worse.
- Each year accounting students from colleges across the nation take the CPA exam. Nationally, the pass rate is one-third. West Chester students traditionally fare very well on these exams. Recently, forty-five West Chester students took the CPA exam and eighteen passed on their first try. Can we be reasonably sure that a higher percentage West Chester students pass the exam?
Proportions:
Ho : π ≤ .33 (wcu is no better than average)
H1 : π > .33 (wcu is better than average)
Since we are trying to show that wcu is better, that has to be the alternative.
Z = (.4 - .33)/√(.33)(.67)/45)=1
.10Reject Ho
.33 1 1.28
We can not be at least 90 percent sure that WCU is better.
the p-value is 16% so we can be 84% sure that WCU is better..