AnswersCA-Statistics and Probability BIntroduction to Confidence Intervals

Introduction to Confidence Intervals Answers

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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.
3

A study was done to estimate the mean commute time for a population of employed adults. A random sample of 25 employed adults in a particular city was taken, and the 95% confidence interval for mean commute time was calculated to be 34.42 ± 8.14 minutes. A councilman is advocating for road repairs and states that, in this city, people with jobs commute an average of 45 minutes to work each day. Does the confidence interval justify the councilman’s statement?

A
No, because 45 minutes is greater than the upper bound of the confidence interval.
B
Yes, because 45 minutes is greater than the upper bound of the confidence interval.
C
No, because the mean commute time from the 25 sampled adults was 34.42 minutes.
D
Yes, because there were almost certainly some people in the sample of 25 adults whose commute was greater than 45 minutes.
4

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.
6

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.
8

A yogurt company claims that it prints a free yogurt coupon under a randomly selected 20% of its lids. A loyal customer purchases 85 yogurt cups, and records whether each was a winner. After consuming all 85 cups, he is disappointed to see that only 12 (14.1%) of his yogurt cups contained coupon codes. He performs a 99% confidence interval for the proportion of yogurt cups containing coupon codes, obtaining (0.044, 0.238). What conclusion can the customer draw about the yogurt company’s claim?

A
The company’s claim may be justified because 0.2 is in the confidence interval.
B
The company’s claim is definitely justified because 0.2 is in the confidence interval.
C
The company’s claim is not justified because the upper endpoint of the confidence interval is 0.238.
D
The company’s claim is not justified because 12 out of 85 (0.141) is far less than 0.2.
10

An inspector inspects large truckloads of potatoes to determine the proportion with blemishes prior to using the potatoes to make potato chips. She intends to compute a 95% confidence interval for this proportion. To do so, she selects a simple random sample of 90 potatoes, and finds 12 with blemishes. The 95% confidence interval is (0.063, 0.204). What is the correct interpretation for this confidence interval?

A
The inspector can be 95% confident that the proportion of potatoes with blemishes is 0.133.
B
In 95% of samples taken, between 6.3% and 20.4% of the sampled potatoes would have blemishes.
C
The inspector can be 95% confident that the interval from 0.063 to 0.204 captures the proportion of all potatoes on the truck with blemishes.
D
There is a 5% probability that the proportion of all potatoes that have blemishes is either below 0.063 or above 0.204.

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