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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,

$$\begin{gathered} n=\frac{{\left(z_{\frac{伪}{2}}\right)}^{2}0.25}{E^2} \\ =\frac{{1.96}^2脳0.25}{{0.03}^2} \\ =1067.11 \\ 鈮�1068 \end{gathered}$$

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.

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,

$$\begin{gathered} \mathit{n}\mathbf{=}\frac{{\left(z_{\frac{伪}{2}}脳蟽\right)}^{\mathbf{2}}}{{\mathbf{E}}^{\mathbf{2}}} \\ =\frac{{1.96}^2脳{47}^2}{5^2} \\ =339.44 \\ 鈮�340 \end{gathered}$$

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.

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.