Practice exam 3 problems, STAT2301
This practice exam has 17 multiple choice and 6 true/false problems. The actual exam will have 25 multiple choice and 5 T/F.
Multiple choice (please select one answer only)
1) Suppose a 95% confidence interval for the true proportion (p) of people who say they approve of Barack Obama as President is (50%, 56%). One of the below statements is correct, please choose it.
(a) The margin of error is plus or minus 3%.
(b) There is 95% chance that p is bigger than 50%.
(c) There is 95% chance that p is between 50% and 56%.
(d) There is 95% chance that p is 53%.
2) A brake light that has been in use a long time is known to have a mean reaction time of 0.8 seconds. A new brake light is tried out on a random sample of 50 drivers. A 95% confidence interval for the mean reaction time for the new light (m) is (0.78, 0.86) seconds. We are interested in seeing if the two lights are different in terms of mean reaction time. Which of the following is a correct statement? (Better means shorter mean reaction time.)
a. We have a statistically significant result and we conclude the new light is better.
b. We do not have a statistically significant result and cannot say if the new light is better or worse.
c. We do not have a statistically significant result and we conclude the new light is better.
d. We do not have a statistically significant result and we conclude the new light is worse.
3) To estimate m, the mean salary of full professors at American colleges and universities, you obtain the salaries of a random sample of 400 full professors. The sample mean is = $75,820, and the sample standard deviation is s = $7300. A 99% confidence interval for m is
a. $75,820 ± $7300.
b. $75,820 ± $958.49.
c. $75,820 ± $940.24.
d. $75,820 ± $47.92.
4) If we want to estimate p, the population proportion of likely voters that believe the economy’s state is the most urgent national concern with 99% confidence and a margin of error no greater than 3%, how many likely voters need to be surveyed? Assume that you have no idea of the value of p.
a. 22
b. 56
c. 716
d. 1844
5) A TV news program conducts a call-in poll about a proposed city ban on smoking in public places. Of the 2467 callers, 1900 were opposed to the ban. Which of the following statements are true with respect to using this sample to estimate p, the proportion of all TV news viewers that favor such a ban on smoking in public places?
a. There is no way this sample can be viewed as an SRS of all TV news viewers, so we can’t use this sample to estimate p.
b. The population is much larger than the sample, so it’s okay to use this sample to estimate p.
c. n is so large that both the counts of both groups are 5 or more, so it’s okay to use this sample to estimate p.
d. There appear to be no violations to any of the assumptions made in using the methods of this chapter. We should now be able to use this sample to estimate p.
Use the following to answer Questions 6–8.
The American Veterinary Medical Association conducted a survey of veterinary clinics to estimate the proportion that do not treat large animals (cows, horses, etc.). The survey was mailed to a random sample of 120 veterinary clinics throughout the country and of these, 88 responded that they do not treat large animals.
6) The standard error SE of the sample proportion of clinics that do not treat large animals, is
a. 0.02.
b. 0.03.
c. 0.04.
d. 0.05.
7) A 95% confidence interval for p, the proportion of veterinary clinics that do treat large animals, is
a. 0.163 to 0.371.
b. 0.188 to 0.346.
c. 0.200 to 0.333.
d. 0.629 to 0.837.
8) A 95% confidence interval for p, the proportion of veterinary clinics that do not treat large animals
a. would be narrower.
b. would be the same length.
c. would be wider.
d. could be wider or narrower.
9) A national health survey of 1472 U.S. adults (presumably selected randomly) during 2010 revealed that 677 had never smoked cigarettes. The 99% confidence interval for the proportion of U.S. adults in 2010 that have never smoked is (0.426, 0.493). From this we may conclude:
a. We will claim the true proportion of never-smokers is between 0.426 and 0.493.
b. We will claim the true proportion of never-smokers is between 42.6% and 49.3%.
c. We think the true proportion of never-smokers is less than 50%.
d. All of the above.
Use the following to answer Questions 10–14.
A candidate for one of Ohio’s two U.S. Senate seats wishes to compare her support among registered voters in the northern half of the state with her support among registered voters in the southern half of the state. A random sample of 2000 registered voters in the northern half of the state is selected, of which 1062 support the candidate. Additionally, a random sample of 2000 registered voters in the southern half of the state is selected, of which 900 support the candidate.
10) We estimate that the proportion of registered adults in the northern half of the state who support this candidate is
a. 0.469.
b. 0.531.
c. 0.546.
d. 0.825.
11) The estimate for the proportion of registered adults in the southern half of the state who support this candidate is
a. 0.183.
b. 0.375.
c. 0.450.
d. 0.825.
12) For purposes of comparing the two proportions, the sampling distribution for the difference in the sample proportions has standard error
a. 0.300.
b. 0.022.
c. 0.016.
d. 0.0002.
13) A 95% confidence interval for the difference in proportion of registered voters that support this candidate between the northern and southern halves of the state is
a. 0.050 to 0.112.
b. 0.035 to 0.127.
c. 0.040 to 0.122.
d. 0.037 to 0.119.
14) The correct conclusion the candidate should draw is as follows:
a. There is no difference between her support in the northern and southern halves.
b. She has greater support in the northern half.
c. She has greater support in the southern half.
d. We can’t make a claim as to which half she has greater support, if any.
15). A radio show conducts a phone-in survey each morning. Listeners are asked to call in with a response to the question of the day. One morning in 2011, listeners were asked if they supported or opposed term limits for members of Congress. Remarkably, 88% of listeners that called in favored term limits. We may safely conclude that
a. there is overwhelming approval for Congressional term limits among Americans generally.
b. it is unlikely that if all Americans were asked their opinion, the results would differ from that obtained in the poll.
c. there is strong evidence that the majority of Americans believe that there should be Congressional term limits.
d. nothing, except that a great majority of those with strong enough feelings on the issue to call in are in favor of Congressional term limits. We cannot generalize any of this survey’s results to any larger population.
16) Which of the following statements is in general true for confidence intervals?
a. A confidence interval must contain the true population value (parameter).
b. A confidence interval will usually not contain the true population value (parameter).
c. We will claim the true population value (parameter) is in the interval, recognizing we could be wrong.
d. A confidence interval must contain zero.
17) If a confidence interval for a difference in proportions (p1-p2) is (-6%, -2%), which of the following is a correct conclusion?
a. We will claim p1 is smaller than p2.
b. We will claim p1 is larger than p2.
c. We will claim p1=p2.
d. We will make no claim about p1 as compared to p2.
True/False
(18) The z* for 95% confidence for a proportion is always 1.96.
(19) Statistical significance means we claim there is a difference.
(20) If a result is known to be statistically significant, then it must be important.
(21) If we are not sure if we have a practically significant result or not then we know we don’t have a statistically significant result.
(22) We need a random sample to use confidence intervals.
(23) A poll of 1000 people from Dallas is more accurate than a poll of 1000 people from the US, assuming both are random samples.