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AP Statistics - CIP: CI for Proportions

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1 Statistical Inference > CI for proportions · Level 3
In a random sample of 1,500 adults with library cards, 525 said that their favorite books to read in bed are mysteries. What is a 95% confidence interval estimate of the proportion of all adults with library cards whose favorite books to read in bed are mysteries?
A
\(0.35 \pm 1.96\sqrt{\dfrac{(0.5)(0.5)}{1500}}\)
B
\(0.35 \pm 1.96\sqrt{\dfrac{(0.5)(0.5)}{1500}}\)
C
\(0.35 \pm 1.96\sqrt{\dfrac{(0.35)(0.65)}{1500}}\)
D
\(0.35 \pm 1.96\sqrt{\dfrac{(0.35)(0.65)}{1500}}\)
E
\(0.35 \pm 1.96\sqrt{1500(0.35)(0.65)}\)
2 Statistical Inference > CI for proportions · Level 3
A 2018 survey of 1,470 adult Americans found that 29 percent of the American adult population would be willing to pay higher taxes if the government offered "Medicare for all." Which of the following best describes what is meant by the poll having a margin of error of ±3 percent?
A
It means that 3 percent of those surveyed refused to participate in the poll.
B
It would not be unexpected for 3 percent of the population to agree readily to the higher taxes.
C
Between 382 and 470 of the 1,470 adults surveyed responded that they would be willing to pay higher taxes in return for "Medicare for all."
D
If a similar survey of 1,470 American adults was taken weekly, a 3 percent change in each week's results would not be unexpected.
E
It is likely that between 26 percent and 32 percent of American adults would be willing to pay higher taxes in return for "Medicare for all."
3 Statistical Inference > CI for proportions · Level 3
The probability that a "walk-on" makes an intercollegiate team at a particular D-1 university is 17%. If one takes a simple random sample (SRS) of walk-ons and constructs a confidence interval estimate of the acceptance rate, which of the following statements is true?
A
The center of the interval would be 17%.
B
The interval would contain 17%.
C
A 99% confidence interval estimate would contain 17%.
D
All of the above are true statements.
E
None of the above are true statements.
4 Statistical Inference > CI for proportions · Level 3
A telephone survey of 750 registered voters showed that 330 favored legalizing medical marijuana. At what confidence level can we say that between 40% and 48% of the electorate favor this legalization?
A
\(P(0.40 < z < 0.48)\)
B
\(P(\dfrac{0.40-0.44}{\sqrt{750(0.44)(0.56)}} < z < \dfrac{0.48-0.44}{\sqrt{750(0.44)(0.56)}})\)
C
\(P(\dfrac{0.40-0.44}{\sqrt{(0.5)(0.5)/750}} < z < \dfrac{0.48-0.44}{\sqrt{(0.5)(0.5)/750}})\)
D
\(P(\dfrac{0.40-0.44}{\sqrt{(0.40)(0.60)/750}} < z < \dfrac{0.48-0.44}{\sqrt{(0.48)(0.52)/750}})\)
E
\(P(\dfrac{0.40-0.44}{\sqrt{(0.44)(0.56)/750}} < z < \dfrac{0.48-0.44}{\sqrt{(0.44)(0.56)/750}})\)
5 Statistical Inference > CI for proportions · Level 3
In a random survey of 530 students, 28 percent said that they had experienced bullying at some time. With what degree of confidence can the pollster say that 28% ± 3% of students have experienced bullying?
A
12.4%
B
28%
C
87.6%
D
93.8%
E
95%
6 Statistical Inference > CI for proportions · Level 3
An online news magazine periodically gives viewers an opportunity to record electronically their agreement or disagreement with some viewpoint or commentary. On one such occasion, 250 out of 800 respondents agreed with a statement that the most practical way of meeting a potential spouse is through online dating sites. The immediate online calculation concluded that 31.25% of the viewers, with a margin of error of ±3.2%, agreed with the statement. The fine print below stated that the calculation was made with 95% confidence. What is the proper conclusion?
A
We are 95% confident that the proportion of viewers who believe that online dating sites are the most practical way of meeting a potential spouse is between 0.280 and 0.345.
B
Without knowing whether both \(n p\) and \(n(1 - p)\) are ≥ 10, the calculation is inappropriate.
C
Without knowing whether or not the 800 respondents represent less than 10% of the entire population of viewers, the calculation is inappropriate.
D
The z-distribution was used when the t-distribution should have been used, so the calculation is inappropriate.
E
The data set was not a simple random sample, so the calculation is inappropriate.
7 Statistical Inference > CI for proportions · Level 3
There are 25,000 high school students in an extended metropolitan region. As each school's students came in to register for classes, guidance counselors were instructed to use a calculator to pick a random number between 1 and 100. If the number 37 was picked, the student was included in a survey. For one of the many survey questions, 86 percent of the students said they couldn't live without instant messaging. Are all conditions met for constructing a confidence interval of the proportion of this region's teenagers who believe they couldn't live without instant messaging?
A
No, there is no guarantee that a representative random sample is chosen.
B
No, \(n p\) and \(n(1 - p)\) are not both greater than 10.
C
No, the sample size is not less than 10% of the population.
D
No, there is no reason to assume that the population has a normal distribution.
E
Yes, all conditions are met and a confidence interval can be constructed.
8 Statistical Inference > CI for proportions · Level 3
For a given large sample size, which of the following gives the smallest margin of error when calculating a confidence interval for a population proportion?
A
91% confidence with \(\hat{p} = 0.18\)
B
93% confidence with \(\hat{p} = 0.18\)
C
95% confidence with \(\hat{p} = 0.18\)
D
91% confidence with \(\hat{p} = 0.34\)
E
93% confidence with \(\hat{p} = 0.34\)
9 Statistical Inference > CI for proportions · Level 3
We are interested in the proportion \(p\) of people who drive for Uber in a large city. Four percent of a simple random sample of 890 people say they drive for Uber. What is the midpoint for a 98% confidence interval estimate of \(p\)?
A
0.01
B
0.49
C
0.5
D
\(p\)
E
None of these are correct.
10 Statistical Inference > CI for proportions · Level 3
If all else is held constant, how does an interval change when constructing a 97% confidence interval rather than a 93% confidence interval?
A
The interval width increases by 4%.
B
The interval width increases by 20%.
C
The interval width increases by 35.8%.
D
The interval width decreases by 20%.
E
This question cannot be answered without knowing the sample size.
11 Statistical Inference > CI for proportions · Level 3
In a random sample of men, a 95% confidence interval for the proportion who say they have ever spent a night in jail is (0.095, 0.115). In a random sample of women, the proportion who say they have ever spent a night in jail is \(\hat{p} = 0.10\). If the two sample sizes are the same, how does a 95% confidence interval for the women's proportion compare to that of the men's?
A
The women's interval has the same width and a lower point estimate.
B
The women's interval is narrower and has a higher point estimate.
C
The women's interval is narrower and has a lower point estimate.
D
The women's interval is wider and has a higher point estimate.
E
The women's interval is wider and has a lower point estimate.
12 Statistical Inference > CI for proportions · Level 3
A restaurant evaluator reports that 22 percent of all restaurant meals include french fries. A 95% confidence interval is given by (0.202, 0.238). Which of the following is a correct interpretation of the confidence level?
A
We are 95 percent confident that the true proportion of all restaurant meals that include french fries is between 0.202 and 0.238.
B
There is a 0.95 probability that the true proportion of all restaurant meals that include french fries is between 0.202 and 0.238.
C
95 percent of all random samples of the same size chosen from the population result in confidence intervals that contain 0.22.
D
95 percent of all random samples of the same size chosen from the population result in confidence intervals that contain the true proportion of all restaurant meals that include french fries.
E
95 percent of all random samples of the same size chosen from the population have sample proportions in the 0.202 to 0.238 interval.
13 Statistical Inference > CI for proportions · Level 3
In a random sample of 800 married couples, 240 reported that the husband did the family laundry. What is a 90% confidence interval estimate of the proportion of all husbands who do the family laundry?
A
\(0.3 \pm 1.645\sqrt{240(0.3)(0.7)}\)
B
\(0.3 \pm 1.645\sqrt{800(0.3)(0.7)}\)
C
\(0.3 \pm 1.96\sqrt{800(0.3)(0.7)}\)
D
\(0.3 \pm 1.645\sqrt{\dfrac{(0.3)(0.7)}{240}}\)
E
\(0.3 \pm 1.645\sqrt{\dfrac{(0.3)(0.7)}{800}}\)
14 Statistical Inference > CI for proportions · Level 3
A pollster plans to obtain a random sample of 250 adults from which he will construct a 99% confidence interval for the proportion of all adults who believe that global warming is caused by human activity. Which of the following statements must be true?
A
The population proportion will be in the confidence interval.
B
The probability that he will construct a confidence interval containing the sample proportion is 0.99.
C
The probability that he will construct a confidence interval containing the population proportion is 0.99.
D
There is a 0.99 probability that the sample proportion will equal the population proportion.
E
None of the above are true statements.
15 Statistical Inference > CI for proportions · Level 3
In a random sample of 1,500 adults, 46 percent answered that they feel the death penalty is applied fairly by the courts. A margin of error of 2 percent is reported. Which of the following statements is appropriate?
A
The sample is large, the sample is random, and the margin of error is relatively small. So it is reasonable to conclude that 46 percent of all adults feel the death penalty is applied fairly by the courts.
B
Although \(n = 1500\) may seem large, it is actually very small in comparison to the number of all adults. So no conclusion can be drawn with any confidence.
C
The sample proportion plus the margin of error is less than 0.50. So there is evidence that less than half of all adults feel the death penalty is applied fairly by the courts.
D
No treatments are applied. So this is not an experiment, and cause and effect cannot be concluded.
E
\(0.02 < 0.05\), so there is sufficient evidence for any reasonable conclusion.
16 Statistical Inference > CI for proportions · Level 3
Factors involved in the creation of a confidence interval include the sample size, level of confidence, and margin of error. Which of the following is incorrect?
A
For a fixed margin of error, larger samples provide greater confidence.
B
For a given confidence level, halving the margin of error requires a sample twice as large.
C
For a given sample size, reducing the margin of error means lower confidence.
D
For a given sample size, higher confidence means a larger margin of error.
E
For a fixed margin of error, smaller samples mean lower confidence.
17 Statistical Inference > CI for proportions · Level 3
In a survey of a random sample of 450 South Carolinians, 75 percent say they enjoy hot pepper sauce on their food. If the margin of error is 2.3 percent, what is the level of confidence?
A
67%
B
74%
C
87%
D
90%
E
93%
18 Statistical Inference > CI for proportions · Level 3
There have been various reports about restaurants and fish markets mislabeling seafood. A consumer advocacy magazine sponsored a study in which 190 pieces of seafood (out of an estimated 12,000 pieces sold each day in the region of interest) were purchased and analyzed. Laboratory results concluded that 22 percent were mislabeled. What is a 97% confidence interval for the proportion of all seafood servings/packages that were mislabeled?
A
\(0.22 \pm 2.17\sqrt{\dfrac{(0.22)(0.78)}{190}}\)
B
\(0.22 \pm 1.88\sqrt{\dfrac{(0.22)(0.78)}{190}}\)
C
\(0.22 \pm 2.17\sqrt{\dfrac{(0.5)(0.5)}{190}}\)
D
\(0.22 \pm 1.88\sqrt{\dfrac{(0.22)(0.78)}{12000}}\)
E
\(0.22 \pm 2.17\sqrt{\dfrac{(0.22)(0.78)}{12000}}\)
19 Statistical Inference > CI for proportions · Level 3
Is climate change causing more severe storms? In a 2017 pre–Hurricane Irma survey, 486 out of 1,080 adults answered in the affirmative. In a 2017 post–Hurricane Irma survey, 546 out of 1,050 answered affirmatively. Establish a 90% confidence interval estimate of the difference (pre minus post-hurricane) between the proportions of adults pre–Hurricane Irma and post–Hurricane Irma who believe climate change is causing more severe storms.
A
\((0.45 - 0.52) \pm 1.645\sqrt{\dfrac{(0.45)(0.55)}{1080} + \dfrac{(0.52)(0.48)}{1050}}\)
B
\((0.45 - 0.52) \pm 1.645(\sqrt{\dfrac{(0.45)(0.55)}{1080}} + \sqrt{\dfrac{(0.52)(0.48)}{1050}})\)
C
\((0.45 - 0.52) \pm 1.96\sqrt{\dfrac{(0.45)(0.55)}{1080} + \dfrac{(0.52)(0.48)}{1050}}\)
D
\((0.45 - 0.52) \pm 1.96(\sqrt{\dfrac{(0.45)(0.55)}{1080}} + \sqrt{\dfrac{(0.52)(0.48)}{1050}})\)
E
\((0.45 - 0.52) \pm 2.576(\sqrt{\dfrac{(0.45)(0.55)}{1080}} + \sqrt{\dfrac{(0.52)(0.48)}{1050}})\)
20 Statistical Inference > CI for proportions · Level 3
In a random sample of 500 high school students, 350 said they had TVs in their bedrooms. In a random sample of 500 middle school students, 325 said they had TVs in their bedrooms. Which of the following represents a 99% confidence interval estimate for the difference (high school minus middle school) between the proportions of high school students and middle school students who have TVs in their bedrooms?
A
\((0.7 - 0.65) \pm 1.96\sqrt{\dfrac{(0.7)(0.3)}{500} + \dfrac{(0.65)(0.35)}{500}}\)
B
\((0.7 - 0.65) \pm 2.326\sqrt{\dfrac{(0.7)(0.3)}{500} + \dfrac{(0.65)(0.35)}{500}}\)
C
\((0.7 - 0.65) \pm 2.576\sqrt{\dfrac{(0.7)(0.3)}{500} + \dfrac{(0.65)(0.35)}{500}}\)
D
\((0.7 - 0.65) \pm 2.326\sqrt{(0.675)(0.325)\left(\dfrac{1}{500} + \dfrac{1}{500}\right)}\)
E
\((0.7 - 0.65) \pm 2.576\sqrt{(0.675)(0.325)\left(\dfrac{1}{500} + \dfrac{1}{500}\right)}\)
21 Statistical Inference > CI for proportions · Level 3
A statistician is developing a method to compare the quotient of proportions from two political polling companies. He tests this method by simulating 5,000 samples from an appropriate population and calculating a 95% confidence interval from each using his new method. If his new method works as intended, 4,750 of the confidence intervals from the simulation will capture what?
A
The difference of sample proportions
B
The quotient of sample proportions
C
The quotient of sample proportions divided by an appropriate standard deviation
D
The difference of population proportions
E
The quotient of population proportions
22 Statistical Inference > CI for proportions · Level 3
In a survey of randomly selected Americans over the age of 65, 40 percent of 756 men and 49 percent of 825 women suffered from some form of arthritis. What is a 96% confidence interval for the difference in the proportions of senior men and senior women who have this disease, and does it indicate sufficient evidence that the proportions are not equal?
A
The confidence interval is (–0.138, –0.042). This does not indicate sufficient evidence that the proportions aren't equal because 0 is not in the interval.
B
The confidence interval is (–0.141, –0.039). This does not indicate sufficient evidence that the proportions aren't equal because 0 is not in the interval.
C
The confidence interval is (–0.138, –0.042). This does indicate sufficient evidence that the proportions aren't equal because 0 is not in the interval.
D
The confidence interval is (–0.141, –0.039). This does indicate sufficient evidence that the proportions aren't equal because 0 is not in the interval.
E
The confidence interval is (–0.138, –0.042). This does not indicate sufficient evidence that the proportions aren't equal because the entire interval is less than 0.

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