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AP Statistics - CIM: CI for Means

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1 Statistical Inference > CI for means · Level 3
Two confidence interval estimates from the same sample are (52.7, 58.3) and (51.8, 59.2). One estimate is at the 95% level, while the other is at the 99% level. Which is which?
A
(51.8, 59.2) is the 95% level.
B
(51.8, 59.2) is the 99% level.
C
This question cannot be answered without knowing the sample size.
D
This question cannot be answered without knowing the sample standard deviation.
E
This question cannot be answered without knowing both the sample size and the sample standard deviation.
2 Statistical Inference > CI for means · Level 3
When making an inference about a population mean, which of the following suggests the use of z-scores rather than t-scores?
A
The absence of outliers
B
The population is normal
C
The sample size is under 30
D
The absence of strong skew
E
The population standard deviation is known
3 Statistical Inference > CI for means · Level 3
Under what conditions would it be meaningful to construct a confidence interval estimate when the data consist of the entire population?
A
If the population size is small \((n < 30)\)
B
If the population size is large \((n \geq 30)\)
C
If a higher level of confidence is desired
D
If the population is truly random
E
Never
4 Statistical Inference > CI for means · Level 3
The number of cloudy days per month (days in which the sun is never seen) in a northwestern town is noted for a random sample of 60 months with \(\overline{x} = 23.4\) and \(s = 3.7\). With what degree of confidence can we assert that the mean number of cloudy days per month in this town is between 22.4 and 24.4?
A
48%
B
90%
C
95%
D
96%
E
99%
5 Statistical Inference > CI for means · Level 3
Four statistics majors receiving actuarial job offers after graduation will receive the following starting salaries: \$54,000, 61,000, \$48,000, and \$57,000. If all assumptions are met, what is a 95% confidence interval for the population mean for all statistics majors receiving actuarial job offers?
A
\(55000 \pm 1.96\left(\dfrac{5477}{\sqrt{3}}\right)\)
B
\(55000 \pm 2.776\left(\dfrac{5477}{\sqrt{3}}\right)\)
C
\(55000 \pm 3.182\left(\dfrac{5477}{\sqrt{3}}\right)\)
D
\(55000 \pm 2.776\left(\dfrac{5477}{\sqrt{4}}\right)\)
E
\(55000 \pm 3.182\left(\dfrac{5477}{\sqrt{4}}\right)\)
6 Statistical Inference > CI for means · Level 3
Two 95% confidence interval estimates are obtained: I (34.5, 40.5) and II (36.3, 44.2). a. If the sample sizes are the same, which has the larger standard deviation? b. If the sample standard deviations are the same, which has the larger sample size?
A
a. I, b. I
B
a. I, b. II
C
a. II, b. I
D
a. II, b. II
E
More information is needed to answer these questions.
7 Statistical Inference > CI for means · Level 3
Suppose (59, 72) is a 99% confidence interval estimate for a population mean \(\mu\). Which of the following is a true statement?
A
Confidence level cannot be interpreted until after data are obtained.
B
There is a 0.99 probability that \(\mu\) is between 59 and 72.
C
The probability that \(\mu\) is in any particular confidence interval can be any value between 0 and 1.
D
If 100 random samples of the given size are picked and a 99% confidence interval estimate is calculated from each, \(\mu\) will be in 99 of the resulting intervals.
E
If 99% confidence intervals are calculated from all possible samples of the given size, \(\mu\) will be in 99% of these intervals.
8 Statistical Inference > CI for means · Level 3
A principal is informed that among the 376 seniors in her high school, the average SAT Math score is 615 with a standard deviation of 28.5. With what margin of error is the mean SAT Math score of these students known?
A
0
B
28.5
C
\(\dfrac{28.5}{\sqrt{615}}\)
D
\(1.96\left(\dfrac{28.5}{\sqrt{615}}\right)\)
E
None of the above give the correct answer.
9 Statistical Inference > CI for means · Level 3
In a random sample of 18 high school students' backpacks, the average weight was 14.5 pounds with a standard deviation of 3.9 pounds. Assuming all conditions for inference are met, what is a 96% confidence interval for the mean weight of all high school students' backpacks?
A
\(14.5 \pm 2.054\dfrac{3.9}{\sqrt{18}}\)
B
\(14.5 \pm 2.205\dfrac{3.9}{\sqrt{18}}\)
C
\(14.5 \pm 2.214\dfrac{3.9}{\sqrt{18}}\)
D
\(14.5 \pm 2.224\dfrac{3.9}{\sqrt{18}}\)
E
\(14.5 \pm 2.235\dfrac{3.9}{\sqrt{18}}\)
10 Statistical Inference > CI for means · Level 3
To determine the average spent on books and lab supplies during a year in college, a simple random sample of 50 students is interviewed, showing a mean of \$970 with a standard deviation of \$310. Which of the following is the best interpretation of a 95% confidence interval estimate for the average spent on books and lab supplies during a year in college?
A
95% of college students spend between \$882 and \$1,058 on books and lab supplies yearly.
B
95% of college students spend a mean dollar amount on books and lab supplies yearly that is between \$882 and \$1,058.
C
We are 95% confident that college students spend between \$882 and \$1,058 on books and lab supplies yearly.
D
We are 95% confident that college students spend a mean dollar amount between \$882 and \$1,058 on books and lab supplies yearly.
E
We are 95% confident that in the chosen sample, the mean dollar amount spent on books and lab supplies yearly by college students is between \$882 and \$1,058.
11 Statistical Inference > CI for means · Level 3
A survey was conducted involving 75 of the 26,000 families living in a town. In 2017, the average amount of medical expenses paid per family in the sample was \$1,950 with a standard deviation of \$640. What is a 95% confidence interval estimate for the total medical expenses paid by all families in the town?
A
\(26000[1950 \pm 1.993(640)]\)
B
\(1950 \pm 1.993\left(\dfrac{640}{\sqrt{75}}\right)\)
C
\(26000[1950 \pm 1.993\left(\dfrac{640}{\sqrt{75}}\right)]\)
D
\(26000[1950 \pm 1.993\left(\dfrac{640}{\sqrt{74}}\right)]\)
E
\(26000[1950 \pm 1.96\left(\dfrac{640}{\sqrt{75}}\right)]\)
12 Statistical Inference > CI for means · Level 3
A home economist wants to estimate the mean temperature of the typical shower. In a random sample of 16 people, the mean temperature of their showers was 101°F with a standard deviation of 2.4°F. Assuming all conditions of inference were met, what is a 95% confidence interval estimate for the mean typical shower temperature for all people?
A
\(101 \pm \dfrac{1.96(2.4)}{4}\)
B
\(101 \pm 1.96\left(\dfrac{\sqrt{2.4}}{4}\right)\)
C
\(101 \pm 2.131\left(\dfrac{\sqrt{2.4}}{16}\right)\)
D
\(101 \pm \dfrac{2.131(2.4)}{4}\)
E
\(101 \pm \dfrac{2.131(2.4)}{16}\)
13 Statistical Inference > CI for means · Level 3
A national education journal states that high school students get an average of 7.6 hours of sleep per night. A guidance counselor thinks that her school's students might have a different average and collects data from a random sample of these students. The data are used to test the hypotheses \(H_0: \mu = 7.6\), \(H_a: \mu \neq 7.6\). The results of the test are shown in the table below.
Sample Mean Standard Error df t P
7.4 0.1091 20 –1.833 0.0817
Assuming all conditions for inference are met, which of the following gives a 90% confidence interval for \(\mu\)?
A
\(7.4 \pm 0.0411\)
B
\(7.4 \pm 0.0436\)
C
\(7.4 \pm 0.1091\)
D
\(7.4 \pm 0.1882\)
E
\(7.4 \pm 0.2000\)
14 Statistical Inference > CI for means · Level 3
A random sample of 60 high school students took part in a study where they ate helpings of pizza and then, four hours later, had their LDL cholesterol levels measured. The 95% confidence interval for the mean LDL cholesterol level for all high school students four hours after eating pizza was calculated to be (103, 120). Which of the following is a true statement?
A
95 percent of all high school students will have LDL cholesterol levels between 103 and 120 four hours after eating pizza.
B
Approximately 95 percent of all random samples of size 60 of high school students will result in 95% confidence intervals containing 111.5 for the mean LDL cholesterol level of all high school students four hours after eating pizza.
C
Approximately 95 percent of all random samples of size 60 of high school students will result in 95% confidence intervals containing the true mean population LDL cholesterol level for the mean LDL cholesterol level of all high school students four hours after eating pizza.
D
The probability is 0.95 that a randomly selected high school student will have an LDL cholesterol level between 103 and 120 four hours after eating pizza.
E
We are 95 percent confident that the mean LDL cholesterol level of these 60 students four hours after eating pizza is between 103 and 120.
15 Statistical Inference > CI for means · Level 3
A 90% confidence interval for the mean time, in minutes, for high school students to finish a video game is determined to be (35.2, 73.4). Which of the following is the best interpretation of the interval?
A
Ten percent of players will take either less than 35.2 minutes or more than 73.4 minutes.
B
The probability that a randomly selected player will take between 35.2 and 73.4 minutes is 0.90.
C
For 90 percent of players, their mean finishing time is between 35.2 and 73.4 minutes.
D
If this study was repeated many times, the resulting confidence interval will contain the true mean finishing time 90 percent of the time.
E
We are 90 percent confident that the true mean finishing time is between 35.2 and 73.4 minutes.
16 Statistical Inference > CI for means · Level 3
Americans and Japanese were surveyed as to the ideal number of children a family should have. For 500 Americans, the mean was 1.95 with a standard deviation of 0.65. For 450 Japanese, the mean was 1.05 with a standard deviation of 0.40. Assuming all conditions for inference are met, what is a 90% confidence interval for the difference in population means?
A
\(0.9 \pm 1.96\sqrt{\dfrac{(1.95)^2}{500} + \dfrac{(1.05)^2}{450}}\)
B
\(0.9 \pm 1.645\left(\dfrac{1.65}{\sqrt{500}} + \dfrac{0.4}{\sqrt{450}}\right)\)
C
\(0.9 \pm 1.645\sqrt{\dfrac{(0.65)^2}{500} + \dfrac{(0.4)^2}{450}}\)
D
\(0.9 \pm 1.96\left(\dfrac{0.65}{\sqrt{500}} + \dfrac{0.4}{\sqrt{450}}\right)\)
E
\(0.9 \pm 3.29\left(\dfrac{0.65}{\sqrt{500}} + \dfrac{0.4}{\sqrt{450}}\right)\)
17 Statistical Inference > CI for means · Level 3
A dietician plans to compare the average number of minutes spent eating each day for French and Americans. The 95% confidence interval estimate of the difference (French – Americans) is (47, 73). Which of the following is the most reasonable conclusion?
A
The mean number of minutes daily spent eating among the French is \(\dfrac{73}{47}\) times the mean spent by Americans.
B
The mean number of minutes daily spent eating among the French is 73 minutes, while the mean among Americans is 47 minutes.
C
The probability that the mean numbers of minutes spent eating are different is 0.95.
D
The probability that the difference in minutes spent eating is greater than 47 minutes is 0.95.
E
We should be 95 percent confident that the difference in mean minutes spent eating is between 47 and 73 minutes.
18 Statistical Inference > CI for means · Level 3
In a national sleep study, a random sample of adults was contacted. These subjects were asked the number of hours they slept per night and whether or not they exercised regularly. Of the 240 adults who exercised regularly, the mean number of hours of sleep per night was 8.2 with a standard deviation of 0.4. Of the 347 adults who did not exercise regularly, the mean number of hours of sleep per night was 7.9 with a standard deviation of 0.5. Which of the following is the most appropriate standard error for a confidence interval of the difference in nightly sleep hours between exercisers and nonexercisers?
A
\(\sqrt{\dfrac{(8.2)^2}{240} + \dfrac{(7.9)^2}{347}}\)
B
\(\dfrac{8.2}{\sqrt{240}} + \dfrac{7.9}{\sqrt{347}}\)
C
\(\sqrt{\dfrac{(0.4)^2}{240} + \dfrac{(0.5)^2}{347}}\)
D
\(\dfrac{0.4}{\sqrt{240}} + \dfrac{0.5}{\sqrt{347}}\)
E
\(\dfrac{\sqrt{(0.4)^2 + (0.5)^2}}{240 + 347}\)
19 Statistical Inference > CI for means · Level 3
A botanist chooses a random sample of 10 maple trees and an independent random sample of 10 elm trees. She measures the diameters of the trees and then calculates the mean and standard deviation of the diameters for each sample. Assume the distribution of diameter sizes for maples and for elms are both normal. What is the appropriate method for constructing a two-sample confidence interval of the difference in the mean diameters of maples and elms?
A
A z-interval is appropriate because both distributions are normal.
B
A z-interval is appropriate because both sample standard deviations are known.
C
A z-interval is appropriate because the central limit theorem applies.
D
A t-interval is appropriate because the population standard deviations are unknown.
E
A t-interval is appropriate because the sample sizes are both below 30.
20 Statistical Inference > CI for means · Level 3
Suppose a two-sample t-test for \(H_0: \mu_1 - \mu_2 = 0\) and \(H_a: \mu_1 - \mu_2 > 0\) results in a P-value of 0.04. Which of the statements below must be true? I. A 90% confidence interval for the difference in means will contain 0. II. A 95% confidence interval for the difference in means will contain 0. III. A 99% confidence interval for the difference in means will contain 0.
A
I only
B
I and II only
C
I, II, and III
D
III only
E
II and III only
21 Statistical Inference > CI for means · Level 3
A nutritionist analyzes two types of hot dogs, all-beef and all-chicken. A random sample of 10 all-beef hot dogs showed a mean of 9.9 mg of fat with a standard deviation of 1.5 mg. A random sample of 8 all-chicken hot dogs showed a mean of 7.5 mg of fat with a standard deviation of 1.1 mg. What is the standard error of the difference in fat content (beef – chicken) between the sample means?
A
\(\sqrt{\dfrac{9.9^2}{10} + \dfrac{7.5^2}{8}}\)
B
\(\dfrac{9.9 - 7.5}{\sqrt{10}\sqrt{8}}\)
C
\(\sqrt{\dfrac{1.5^2 - 1.1^2}{10 \cdot 8}}\)
D
\(\sqrt{\dfrac{1.5^2}{10} + \dfrac{1.1^2}{8}}\)
E
\(\dfrac{1.5}{\sqrt{10}} + \dfrac{1.1}{\sqrt{8}}\)
22 Statistical Inference > CI for means · Level 3
An experiment is conducted to test the effectiveness of a new sleep aid. Twenty volunteers are randomly assigned to two groups, with one group receiving the sleep aid while the other group receives a similar-looking placebo. The next day, the mean hours of sleep for each group is calculated and a two-sample t-test for the difference of means is performed. Which of the following is a necessary assumption?
A
The expected number of hours of sleep for each person is at least 5.
B
The sample sizes are greater than or equal to 10 percent of the population sizes.
C
The 10 volunteers randomly assigned to the sleep aid group are paired with the 10 volunteers assigned to the placebo group.
D
The distributions of hours of sleep in each population is approximately normal.
E
Each group of 10 should have 5 successes and 5 failures.

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