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Chapter 8: Q. R8.6 (page 548)

Do you go to church? The Gallup Poll plans to ask a random sample of adults

whether they attended a religious service in the last 7 days. How large a sample would be required to obtain a margin of error of at most 0.01 in a 99% confidence interval for the population proportion who would say that they attended a religious service?

Short Answer

Expert verified

The required sample size is3152

Step by step solution

01

Given information

Confidence level=99%

Margin of error (E)=0.01

02

The objective is to find out the sample size to have a margin of error of at most 0.01at 99%a confidence level.

We know,

The formula to compute the sample size is:

n=p^(1-p^)zα2E2

There is no information about p^provided here.

Hence, assume is to be 0.50.

According to the standard normal table, the value of za2corresponding to the 99%confidence level is2.576

The sample size can be calculated as:

localid="1654402152379" n=p^(1-p^)zα2E2=0.50(1-0.50)(2.5760.01)2=3151.576≈3152

Therefore, the sample size is3152

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Most popular questions from this chapter

A plethora of pepperoni? Refer to Exercise 73.

a. Explain why it was necessary to inspect a graph of the sample data when checking the Normal/Large Sample condition.

b. According to the manager of the restaurant, there should be an average of 40

pepperonis on a large pizza. Based on the interval, is there convincing evidence that the average number of pepperonis is less than 40? Explain your answer.

Selling online According to a recent Pew Research Center report, many

American adults have made money by selling something online. In a random sample of 4579American adults, 914reported that they earned money by selling something online in the previous year. Assume the conditions for inference are met.

Determine the critical value z* for a 98% confidence interval for a proportion.

b. Construct a 98% confidence interval for the proportion of all American adults who

would report having earned money by selling something online in the previous year.

c. Interpret the interval from part (b).

Bone loss by nursing mothers Breast-feeding mothers secrete calcium into

their milk. Some of the calcium may come from their bones, so mothers may lose bone mineral. Researchers measured the percent change in bone mineral content (BMC) of the spines of 47randomly selected mothers during three months of breast-feeding. The mean change in BMC was −3.587%and the standard deviation was 2.506%

a. Construct and interpret a 99%confidence interval to estimate the mean percent change in BMC in the population of breast-feeding mothers.

b. Based on your interval from part (a), do these data give convincing evidence that

nursing mothers lose bone mineral, on average? Explain your answer.

Suppose we want a 90% confidence interval for the average amount spent on books by freshmen in their first year at a major university. The interval is to have a margin of error of at most \(2. Based on last year’s book sales, we estimate that the standard deviation of the amount spent will be close to \)30. The number of observations required is closest to which of the following?

a. 25

b.30

c. 225

d. 609

e. 865

How confident? The figure shows the result of taking 25SRSs from a Normal population and constructing a confidence interval for the population mean using each sample. Which confidence level—80%,90%,95%,or99%—do you think was used? Explain your reasoning.
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