Cumulative Exam — Cumulative exam Answers

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23

An animal rescue agent wanted to estimate the true proportion of all animals in shelters that are adopted each month. To do so, she selects a random sample of 100 animals and determines that the 95% confidence interval for the true proportion of animals adopted each month is between 0.12 and 0.24. Which of these statements is a correct interpretation of the confidence level?

A
Approximately 95% of all animals in the shelter were adopted.
B
There is a 95% probability that the true proportion of all animals in the shelter that were adopted last month is between 0.12 and 0.24.
C
If many random samples of size 100 are selected from all records of animals in shelters, approximately 95% of the intervals would capture the true proportion that were adopted.
D
Approximately 95% of the sample proportions, based on random samples of size 100 from the population of all records of animals in shelters, will fall between 0.12 and 0.24.
26

A teacher has two large containers filled with blue, red, and green beads, and claims the proportions of red beads are the same in each container. Each student shakes the first container, selects 50 beads, counts the number of red beads, and returns the beads to the container. The student repeats this process for the second container. One student’s samples contained 13 red beads from the first container and 16 red beads from the second container.Based on the 95% confidence interval, (–0.24, 0.11), is the teacher’s claim justified?

A
The teacher’s claim is not justified because the two sample proportions are different.
B
The teacher’s claim is not justified because the interval contains both positive and negative values.
C
The teacher’s claim is justified because the interval contains 0, meaning there is no difference in the proportions of beads in each container.
D
The teacher’s claim might not be justified because the interval contains both positive and negative values. It is plausible that the proportion of red beads in each container is different.
28

In a small town of 5,832 people, the mayor wants to determine if there is a difference in the proportion of voters ages 18–30 who would support an increase in the food tax, and the proportion of voters ages 31–40 who would support an increase in the food tax. An assistant to the mayor surveys 85 randomly chosen voters ages 18–30, and finds that 62 support the increase. A random sample of 70 voters ages 31–40 is also surveyed, and 56 support the increase. Assuming the conditions for inference have been met, what is the 99% confidence interval for the difference in proportions of voters who would support the increase in the food tax for the different age groups?Find the z-table here.

A
(0.27 minus 0.20) plus-or-minus 1.96 StartRoot StartFraction 0.27 (1 minus 0.27) Over 85 EndFraction + StartFraction 0.20 (1 minus 0.20) Over 70 EndFraction EndRoot
Option A
B
(0.73 minus 0.80) plus-or-minus 1.96 StartRoot StartFraction 0.73 (1 minus 0.73) Over 70 EndFraction + StartFraction 0.80 (1 minus 0.80) Over 85 EndFraction EndRoot
Option B
C
(0.27 minus 0.20) plus-or-minus 2.58 StartRoot StartFraction 0.27 (1 minus 0.27) Over 70 EndFraction + StartFraction 0.20 (1 minus 0.20) Over 85 EndFraction EndRoot
Option C
D
(0.73 minus 0.80) plus-or-minus 2.58 StartRoot StartFraction 0.73 (1 minus 0.73) Over 85 EndFraction + StartFraction 0.80 (1 minus 0.80) Over 70 EndFraction EndRoot
Option D
29

A statistics class weighed 20 bags of grapes purchased from the store. The bags are advertised to contain 16 ounces, on average. The class calculated the 90% confidence interval for the true mean weight of bags of grapes from this store to be (15.875, 16.595) ounces. Is the store justified in stating that the average weight of the bags of grapes is 16 ounces?

A
The store is not justified in stating that the average weight of the bags is 16 ounces because the sample mean of 16.235 ounces is more than 16 ounces.
B
The store may be justified in stating that the average weight of the bags is 16 ounces because 16 ounces is in the confidence interval.
C
The store is justified in stating that the average weight of the bags is 16 ounces because 16 ounces is in the confidence interval.
D
The store is not justified in stating that the average weight of the bags is 16 ounces because the majority of the confidence interval is above 16 ounces.
31

A school principal claims the graduation rate at a school is 96%. Molly, a student at this school, takes a random sample of students and finds the 95% confidence interval for the true proportion of students graduating from this school is (0.934, 0.983). Is it reasonable to conclude the principal’s claim is incorrect?

A
No, because the interval contains 0.96.
B
No, because the interval contains values greater than 0.96.
C
Yes, because the interval has values less than 0.96.
D
Yes, because a different sample might give different results.
33

A college performs a survey of 424 randomly chosen graduates to estimate the proportion of alumni who are working in the field of their college degree. For example, if a student earned a degree in biology, do they work in the field of biology? Of the 424 alumni, 361 reported that they were working in the field of their college degree. A 98% confidence interval for the true proportion of graduates who are working in the field of their degree is (0.811, 0.892). What is the correct interpretation of the confidence interval?

A
98% of all polls of graduates like these will give sample proportions in the range from 0.811 to 0.892.
B
The percentage of all graduates who are working in the field of their degree is certainly between 0.811 and 0.892.
C
It can be stated with 98% confidence that the proportion of all graduates who are working in the field of their degree is captured by the interval from 0.811 to 0.892.
D
It can be stated with 98% confidence that the sample proportion of graduates who are working in the field of their degree is contained within the confidence interval.
34

In a statistics activity, students are asked to determine if there is a difference in the proportion of times that a spinning penny will land with tails up, and the proportion of times a spinning dime will land tails up. The students are instructed to spin the penny and the dime 30 times and record the number of times they land tails up. For one student, the penny lands tails side up 18 times, and the dime lands tails side up 20 times. Assuming the conditions for inference are met, what is the 98% confidence interval for the difference in proportions of tails side up for a penny and a dime?Find the z-table here.

A
(0.40 minus 0.33) plus-or-minus 2.33 StartRoot StartFraction 0.40 (1 minus 0.40) Over 30 EndFraction + StartFraction 0.33 (1 minus 0.33) Over 30 EndFraction EndRoot
Option A
B
(0.40 minus 0.33) plus-or-minus 2.58 StartRoot StartFraction 0.40 (1 minus 0.40) Over 30 EndFraction + StartFraction 0.33 (1 minus 0.33) Over 30 EndFraction EndRoot
Option B
C
(0.60 minus 0.67) plus-or-minus 2.33 StartRoot StartFraction 0.60 (1 minus 0.60) Over 30 EndFraction + StartFraction 0.67 (1 minus 0.67) Over 30 EndFraction EndRoot
Option C
D
(0.60 minus 0.67) plus-or-minus 2.58 StartRoot StartFraction 0.60 (1 minus 0.60) Over 30 EndFraction + StartFraction 0.67 (1 minus 0.67) Over 30 EndFraction EndRoot
Option D
35

The owner of a popular coffee shop believes that customers who drink coffee are more likely to use their own cup than customers who drink espresso. Customers using their own cups get a 5% discount, which is displayed on the receipt. The owner randomly selects 50 receipts from all coffee purchases and 50 receipts from all espresso purchases. For coffee purchases, 24 receipts showed that the customer used their own cup. For espresso purchases, 18 receipts showed the customer used their own cup.Based on the 99% confidence interval, (–0.13, 0.37), is the coffee shop owner’s claim justified?

A
The owner’s claim is justified because there are positive values for the proportion of customers who use their own cups.
B
The owner’s claim is justified because the interval contains both positive and negative values for the difference in the proportions.
C
The owner’s claim is not justified because the proportion of customers who purchase coffee beverages and use their own cups is different from the proportion of customers who purchase espresso beverages.
D
The owner’s claim is not justified because the interval contains nonpositive values. It is plausible that there is no difference, or that the proportion of espresso customers using their own cups is higher than the proportion of coffee customers using their own cups.
36

In a statistics activity, students are asked to determine if there is a difference in the proportion of times that a spinning penny will land with tails up, and the proportion of times a spinning dime will land tails up. The students are instructed to spin the penny and the dime 30 times and record the number of times they land tails up. For one student, the penny lands tails side up 18 times, and the dime lands tails side up 20 times.Based on the 98% confidence interval, (–0.36, 0.22), is there evidence of a difference in proportions of tails side up for a penny and a dime?

A
There is convincing evidence because the two sample proportions 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 –0.07. Since this is not 0, there is a difference in the true proportions of tails up for pennies and dimes.
D
There is not convincing evidence because the interval contains both negative and positive values for the true difference.
38

A newspaper poll found that 54% of the respondents in a random sample of voters in the city plan to vote for candidate Roberts. A 95 percent confidence interval for the population proportion is 0.54 ± 0.06. Based on this interval, what can the newspaper report?

A
There is a 95% chance that Roberts will win.
B
The race is too close to call, so no prediction about who will win should be made.
C
With 95% confidence, there is convincing evidence that Roberts will win.
D
The poll predicts Roberts will win, but there is a 5% chance that the prediction is incorrect due to sampling error.
39

A researcher for a polling organization used a random sample of 1,540 residents in a city to construct a 95 percent confidence interval for the proportion of voters who would vote for candidate Jones. The resulting confidence interval was 0.480 ± 0.025. What is the correct interpretation of the confidence interval?

A
A proportion of 0.455 and 0.505 of respondents think that Jones has a 95% chance to win.
B
If 95% of all the voters voted, then Jones would receive between 45.5% and 50.5% of the votes.
C
The polling organization can be 95% confident that the interval from 0.455 to 0.505 captures the proportion of all city voters who would vote for Jones.
D
If we repeatedly sampled voters from this city, taking samples of size 1,540, approximately 95% of those samples would have between 45.5% and 50.5% voting for Jones.

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