Practice Problems for Confidence Intervals
X ± Z σ/n ; X ± t s/n ; p ± Z (p(1-p)/n)
1. A press release issued by our university claims that West Chester students study at least as much as
the national average for students at four year universities. Across the nation, 73 percent of all
students at four year universities study at least four hours per week. Seventy percent of one
hundred randomly selected West Chester students surveyed claimed to study more than four hours
per week. Should the University retract its previous statement? Explain why or why not.
We can be 90 percent sure that the percentage of all WCU students who study at least 4 hours per
week is :
.7  1.64 (.7*.3/100)
.7  .07. so 63 to 77%. We can’t be sure the statement is correct, but it might be so I would not
instruct the University to retract its statement.
2. A car manufacturer is considering switching to a new brand of brake shoes. To test the shoes, the
manufacturer submits the shoes to a heat test. The higher the failure temperature, the better the
shoe. Past experience has shown that heat failure temperature is normally distributed. A sample of
five shoes failed at the following temperatures: 300, 310, 280, 350 and 260F. If the old shoes,
which are considered reliable fail at an average temperature of 290F, should the company make the
switch? Explain your answer.
Using the t distribution, we can be 95 percent sure that the average failure temperature of the new
shoes is: 300  2.776(33.91/5) (that t is based on .025 and d.f. = 4)
= 300  42 258 to 342. We cannot be sure that the new pads are better (though they may be) sow we
should not switch)
3. A government researcher is trying to determine whether or not a new car model averages at least 30
miles per gallon in the city. If it does, the government will give the manufacturer a $100 tax break
per car. From previous studies, the standard deviation of mpg is known to be 5 mpg. If the
average mpg of the new model in a sample of 36 is 29mpg, should the government award the tax
break? Explain why or why not.
We can be 90 percent sure that the actual gas mileage is 29  1.28 5/36
29  1.06 27.94 to 30.06 mpg. We can not be certain that the care deserves a tax break, but it might.
If the burden of proof lies with the manufacturer then, no. If the government must prove the
manufacturer has not met the standard then they get the break.
4. A company is thinking about offering a lifetime guarantee for its mufflers. The company figures
that so long as its customers own their cars on average for less than ten years the offer will be
profitable. (The warranty would only apply to the original customer.) The following figures were
collected from a random sample of past customers:
years until sale: 2, 3, 6, 9, 7, 20, 2
Assuming the company is extremely conservative when it comes to trying new ideas, what should the
company do and why? Give the CEO a suggestion for improving the study, and explain why your
suggestion will help.
The agerage length of ownersip (with 99% confidence) is: : 7  3.707(6.32/7)
78.8 = 0 to 15.5. We cannot be sure that the average is less than 10. To improve the study they
should use a sample over 30. This will decrease the width of the interval, and increase the reliability
of the conclusion becasue we will not have to assume that the population is normal.
5. In a simple random sample of 36 West Chester adults, 20 favored some type of tax reform.
Assuming there is little importance to the decision, is it reasonable to assume that the majority of
all West Chester Adults is in favor of tax reform? Explain why or why not.
20/36 is 0.56; as there is little consequence to the decision, use a low 80 confidence level, for which
Z= 1.28
P  Z*P*(1-P)/n = 0.56  1.28*{0.56*0.44/36}
=
0.56  .106 or 45 to 56%
There is a possibility that the majority does not support tax reform, but, of course, they would be
idiots. But I would not assume that tax reform is favored by the majority.
6. A company in the U.S. is considering investing in a new small company in another country. One
executive is concerned about bankruptcy rates in that country. Unfortunately, the country does not
keep accurate bankruptcy rates. However, companies that exist are required to register with the
government each year. A person with some time can follow the companies until they disappear
from the lists. Usually this means they went bankrupt. The company does not have time to do this
for every firm that ever existed, so they try a sample. Of 36 randomly selected firms they followed
through the lists, 4 were bankrupt within 10 years. In the U.S. approximately 8 percent of all
companies go bankrupt within 10 years. Does this indicate that the risk of bankruptcy is higher
than it is in the U.S.? Explain your answer.
4 of 36 companies is 11%, so the question is how confident are we that the average for all companies
in this banana republic is higher than 8
P  Z * {P*(1-P)/n} or 0.11  1.28 * {.11*.89/36} or 0.11  .067
We can be 90 percent sure that the bankruptcy rate for all firms is between 4 and 18 percent. The risk
of bankruptcy may well be higher than the U.S. Bob indicates that we can be 73 percent confident that
the risk is higher. By reducing the confidence level, the interval shrinks at 73% the interval is entirely
above 8 percent. Usually it is better to select the minimum confidence level first and live with the
result.
.
7. Across the nation it takes an average of 4 minutes to process a customer at a drive through fast
food window. In addition, drive through times tend to be normally distributed. At a local
McDunalds the manager thinks his employees are not operating up to national standards. Because
he doesn’t have time to check the speed of every clerk for every customer, it’s difficult to tell for
sure. Based on a suggestion from a colleague, he decides to calculate a confidence interval. From
behind the one way mirror in the booth above the clerks, he times several sales with the following
results:
Customer
Time (in minutes)
Average
Std Dev
1
6.00
6.00
1.62
2
5.50
3
8.50
4
7.00
5
3.00
6
7.00
7
5.00
8
6.00
Assuming the manager uses a low confidence level, what is the manager likely to conclude after his
study? Based on the same information, do you think his employees are not up to national standards?
Explain your answer.
If the manager uses a low confidence level (let’s presume 80%) degrees of freedom = 7, then t = 1.415.
Substituting, we get
6.00  1.415* 1.62/8 =
6.00  .81,
The manager should expect his crew to be between 5.19 and 6.81 minutes. This is greater than the
known national average of 4 minutes. This manager should be concerned about his team’s
performance.
Note: small sample is o.k. in this case because the population is normal.
8. A local company searching for ways to save more money is considering suggesting that its
employees use a “Dial around“ service like “10-10-321, 10-10-220 or Lucky dog to cut down on
the company’s phone bills. The rules and charges for these services vary and are often
complicated, so depending on the length of the call or the time of day, the service may or may not
save money. Nationally the variation in the amount saved per call is around 15 cents. As a test,
the company makes 60 calls with one of the services and finds the average savings on those calls
was 25 cents. While the initial result is encouraging, can we be reasonably sure the service will
save us money? Explain why or why not
There is insufficient information on which to base a decision. We know the variability of the national
data, but not the average national savings per call. Likewise, we know the sample average, but have no
indication of the sample variability. AM I MISSING SOMETHING? Nationally the variation in the
amount saved per call is around 15 cents. I snuck that standard deviation right in there!
.25  1.64 *.15/60
.25  .03 .
We can be 90 percent sure that the average savings will be between .22 and .28. Since we can be sure
we will save ( saving > 0) , the dial around works.
9. "More Americans Refuse to Pay Taxes" read the headline in the April 16th edition of the
Philadelphia Examiner. Last year, 5.2 percent of Americans failed to file an income tax form.
This year in a survey of 67 taxpayers, 6 percent had failed to file. Was the Examiner's headline
accurate or misleading? Explain your reasoning
There would seem to be little damage if the statement were incorrect, so use an 80% confidence level.
Examine the variability of the sample data. z = 1.28 Substituting, we get
.06  1.28*{.06*.94/67}
or
.06  1.28 * .03
or
.06  .037
At an 80% confidence, we would expect our sample to describe a population from 2.3 to 9.7% failing
to file. We have good reason to believe our data, as the sample is large enough to behave normally.
Since last year’s value 5.2% is not outside this range, it would be imprudent (and misleading) to make this
statement.