Question

Sample Size You have been hired by your new employer to survey adults about printednewspaper subscriptions.

a. If you want to estimate the percentage of adults who have a paid subscription to a printednewspaper, how many adults must you survey if you want 95% confidence that your percentagehas a margin of error of three percentage points?

b. If you want to estimate the mean amount that adults have spent on printed newspapers withinthe past year, how many adults must you survey if you want 95% confidence that your samplemean is in error by no more than \(5? (Based on results from a pilot study, assume that the standarddeviation of amounts spent on printed newspapers in the last year is \)47.)

c. If you plan to obtain the estimates described in parts (a) and (b) with a single survey having several questions, how many adults must be surveyed?

Short answer

a. The number of adults that must be surveyedto estimate the percentage of adults who have a paid subscription to a printed newspaper is 1068.

b. The number of adults that must be surveyed to estimate the mean amount that adults have spent on printed newspapers within the past year is 340.

c. The adults that must be surveyed to obtain the estimates described in (a) and (b) parts is 1068.

Step-by-step solution

Compute the sample size of adults who have a paid subscription to a printed newspaper

a.

Given: The level of confidence is 95%. The margin of error is E=0.03.

Since the level of confidence is 95%, this implies the level of significance is 0.05.

From the Z table, the critical value at 0.05 level of significance is 1.96.

The number of adults that must be surveyedto estimate the percentage of adults who have a paid subscription to a printed newspaper is computed as,

$$\begin{gathered} n=\frac{{\left(z_{\frac{\alpha }{2}}\right)}^{2}0.25}{E^2} \\ =\frac{{1.96}^2\times 0.25}{{0.03}^2} \\ =1067.11 \\ \approx 1068 \end{gathered}$$

Thus, the number of adults that must be surveyedto estimate the percentage of adults who have a paid subscription to a printed newspaper is 1068.

Compute the sample size of adults who have spent on printed newspapers within the past year

b.

Given: The level of confidence is 95%.The margin of error is E=$5. The standard deviation of amounts spent on printed newspapers in the last year is $47.

Since the level of confidence is 95%, this implies the level of significance is 0.05.

From the Z table, the critical value at 0.05 level of significance is 1.96.

The number of adults that must be surveyed to estimate the mean amount that adults have spent on printed newspapers within the past year is given as,

$$\begin{gathered} \mathit{n}\mathbf{=}\frac{{\left(z_{\frac{\alpha }{2}}\times \sigma \right)}^{\mathbf{2}}}{{\mathbf{E}}^{\mathbf{2}}} \\ =\frac{{1.96}^2\times {47}^2}{5^2} \\ =339.44 \\ \approx 340 \end{gathered}$$

Therefore, the number of adults that must be surveyed to estimate the mean amount that adults have spent on printed newspapers within the past year is 340.

Compute the sample size

c.

On comparing the above two sample results, the sample size that should be used for single a survey is 1068 as it covers the maximum number of subjects.

Thus, the adults that must be surveyed to obtain the estimates described in (a) and (b) parts is 1068.