Unit Test — Unit test Answers

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3

A student wants to survey the sophomore class of 200 students about whether the school should require uniforms. A random sample of 50 sophomores is surveyed and asked whether they support the school adopting uniforms. Of the 50 sophomores, 12 say they would favor school uniforms. Assuming the conditions for inference have been met, what is the 90% confidence interval for the true proportion of sophomores who favor the adoption of uniforms?

A
0.24 plus-or-minus 1.96 StartRoot StartFraction 0.24 (1 minus 0.24) Over 200 EndFraction EndRoot
Option A
B
0.76 plus-or-minus 1.96 StartRoot StartFraction 0.76 (1 minus 0.24) Over 200 EndFraction EndRoot
Option B
C
0.24 plus-or-minus 1.65 StartRoot StartFraction 0.24 (1 minus 0.24) Over 50 EndFraction EndRoot
Option C
D
0.76 plus-or-minus 1.95 StartRoot StartFraction 0.76 (1 minus 0.76) Over 50 EndFraction EndRoot
Option D
5

A computer company wants to determine if there is difference in the proportions of defective computer chips in a day’s production from two different production plants, A and B. A quality control specialist takes a random sample of 100 chips from the day’s production from plant A, and determines that there are 12 defective chips. The specialist then takes a random sample of 100 chips from the day’s production from plant B, and determines that there are 10 defective chips.Based on the 90% confidence interval, (–0.05, 0.09), is there convincing evidence of a difference in the true proportions of defective chips from a day’s production between the two plants?

A
There is convincing evidence because the sample proportions of defective chips for the two plants are different.
B
There is not convincing evidence because the interval contains 0.
C
There is convincing evidence because the difference in the two sample proportions is 0.02. Since this is not 0, there is a difference in the true proportions of defective chips for the two plants.
D
There is not convincing evidence because the interval contains both negative and positive values.
6

A political candidate feels that she performed particularly well in the most recent debate against her opponent. Her campaign manager polled a random sample of 400 likely voters before the debate and a random sample of 500 likely voters after the debate. The 95% confidence interval for the true difference (post-debate minus pre-debate) in proportions of likely voters who would vote for this candidate was (–0.014, 0.064). What is the correct interpretation of the 95 percent confidence interval?

A
Only 95% of the voters who were polled responded to the question.
B
The candidate can be 95% confident that the percentage of likely voters who would vote for her increased, since much of the confidence interval is above 0.
C
The candidate can be 95% confident that the percentage of likely voters who would vote for her did not increase, since the confidence interval contains 0.
D
We are 95% confident that the interval from –0.014 to 0.064 captures the true change in the proportion of likely voters who would vote for this candidate.
7

A principal of a large high school wants to estimate the true proportion of high school students who use the community’s public library. To do so, he selects a random sample of 50 students and asks them if they use the community’s public library. The 95% confidence interval for the true proportion of all students who use the community’s public library is 0.25 to 0.34. Which of these statements is a correct interpretation of the confidence level?

A
Approximately 95% of the 50 randomly selected students use the community’s public library.
B
The principal can be 95% confident that the interval from 0.25 to 0.34 captures the true proportion of high school students who use the community’s public library.
C
If many random samples of size 50 are selected from all students at this school, approximately 95% of the intervals would capture the true proportion of students who use the community’s public library.
D
Approximately 95% of the intervals, based upon random samples of size 50 from all students in this high school, will find that between 25% and 34% of students at this school use the community’s public library.
10

A statistics student from a large high school takes a random sample of 200 students and finds that 123 are actively involved in a political party. A 99% confidence interval for the proportion of students at this school who are actively involved in a political party is (0.526, 0.704). Which statement correctly interprets the interval?

A
The statistics student can be 99% confident that 62% of students from this school are actively involved in a political party.
B
There is a 99% probability that the proportion of students at this school who are actively involved in a political party lies between 0.526 and 0.704.
C
If this procedure is repeated many, many times, the true proportion of students at this school who are actively involved in a political party will be captured 99% of the time.
D
The statistics student can be 99% confident that the interval from 0.526 to 0.704 captures the true proportion of all students at this school who are actively involved in a political party.
14

A local dentist is concerned that less than half of her patients floss daily. A 95% confidence interval for the true proportion of her patients who floss daily is (0.325, 0.701). Is it reasonable to believe that less than half of her patients floss daily?

A
Yes, because 0.50 is in the interval.
B
Yes, because the majority of the interval is less than 0.50.
C
No, because there are values in the interval greater than 0.50.
D
No, because the interval has a lower bound of 0.325, which is not statistically lower than 0.50.
16

A politician claims that a proposal for a new traffic law is broadly supported by both political parties and that a person from either political party is equally likely to support the proposed legislation. He cites two recent polls that said 70% of a random sample of 550 people from his political party supports the law, and 65% of a random sample of 420 people from the other political party supports the law. The 95 percent confidence interval for the difference in population proportions is (–0.010, 0.110). Based on the interval, is the politician’s claim justified?

A
The politician’s claim is not justified because the two sample proportions are different.
B
The politician’s claim is justified because the interval contains 0, which indicates no difference in the population proportions.
C
The politician’s claim may be justified because the interval contains 0, which indicates no difference in the population proportions. However, because the interval also contains positive and negative values, it is also plausible that there is a difference in the proportions of party members who support the new traffic law.
D
The politician’s claim is not justified because the interval contains both positive and negative numbers.
17

A local school board wants to estimate the difference in the proportion of households with school-aged children that would support starting the school year a week earlier, and the proportion of households without school-aged children that would support starting the school year a week earlier. They survey a random sample of 40 households with school-aged children about whether they would support starting the school year a week earlier, and 30 households respond yes. They survey a random sample of 45 households that do not have school-aged children, and 25 respond yes. Assuming the conditions for inference have been met, what is the 90% confidence interval for the difference in proportions of households that would support starting the school year a week earlier?Find the z-table here.

A
(0.75 minus 0.56) plus-or-minus 1.65 StartRoot StartFraction 0.75 (1 minus 0.75) Over 45 EndFraction + StartFraction 0.56 (1 minus 0.56) Over 40 EndFraction EndRoot
Option A
B
(0.75 minus 0.56) plus-or-minus 1.96 StartRoot StartFraction 0.75 (1 minus 0.75) Over 85 EndFraction + StartFraction 0.56 (1 minus 0.56) Over 85 EndFraction EndRoot
Option B
C
(0.75 minus 0.56) plus-or-minus 1.65 StartRoot StartFraction 0.75 (1 minus 0.75) Over 40 EndFraction + StartFraction 0.56 (1 minus 0.56) Over 45 EndFraction EndRoot
Option C
D
(0.75 minus 0.56) plus-or-minus 1.96 StartRoot StartFraction 0.75 (1 minus 0.75) Over 40 EndFraction + StartFraction 0.56 (1 minus 0.56) Over 45 EndFraction EndRoot
Option D

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