Question

TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streaming is gaining in popularity. The Harris poll reported on November 13, 2012, that 53% of 2343 American adults surveyed said they have watched digitally streamed TV programming on some type of device.

a. Calculate and interpret a confidence interval at the 99% confidence level for the proportion of all adult Americans who watched streamed programming up to that point in time.

b. What sample size would be required for the width of a 99% CI to be at most .05 irrespective of the value of p ˆ?

Short answer

a) The boundaries of the confidence interval are then \((0.5034,0.5566)\)

b) The sample size is \({\text{n - 2653}}\)

Step-by-step solution

Large-sample confidence interval

(a) \(\begin{aligned}n &= 2343 \hfill \\\hat p &= 53\% = 0.53 \hfill \\c &= 99\% = 0.99 \hfill \\ \end{aligned} \)

For confidence level \({\text{1 - \alpha - 0}}{\text{.99}}\) , determine \({{\text{z}}_{{\text{\alpha /2}}}}{\text{ - }}{{\text{z}}_{{\text{0}}{\text{.005}}}}\) using the normal probability table in the appendix (look up 0.005 in the table, the z-score is then the found z-score with opposite sign):

\({{\text{z}}_{{\text{\alpha /2}}}}{\text{ - 2}}{\text{.575}}\)

The margin of error is then:

\({\text{E - }}{{\text{z}}_{{\text{\alpha /2}}}}{\text{ \times }}\sqrt {\frac{{{\text{\hat p(1 - \hat p)}}}}{{\text{n}}}} {\text{ - 2}}{\text{.575 \times }}\sqrt {\frac{{{\text{0}}{\text{.53(1 - 0}}{\text{.53)}}}}{{{\text{2343}}}}} {\text{\gg 0}}{\text{.0266}}\)

The boundaries of the confidence interval are then:

\(\begin{aligned}\hat p - E - 0.53 - 0.0266 - 0.5034 \hfill \\\hat p + E - 0.53 + 0.0266 - 0.5566 \hfill \\\end{aligned} \)

Hence the boundaries of the confidence interval are then \((0.5034,0.5566)\)

Step 2:To find the sample size

(b) Given:

\(\begin{aligned}w&=0.05\hfill\\c&=99\%=0.99\hfill\\\hat p&= 53\%= 0.53 \hfill \\\end{aligned} \)

Formula sample size:

\(\begin{aligned}n\gg &\frac{{4{{\left[ {{z_{\alpha /2}}} \right]}^2}\hat p\hat q}}{{{w^2}}} -\frac{{4{{\left[ {{z_{\alpha /2}}} \right]}^2}\hat p(1 - \hat p)}}{{{w^2}}} - \frac{{4{{\left[ {{z_{\alpha /2}}}\right]}^2}\hat p(1 - \hat p)}}{{{w^2}}} \hfill \\{z_{\alpha /2}} - 2.575 \hfill \\\end{aligned} \)

Note: We take the average of 2.57 and 2.58, because 0.005 is exactly in the middle between 0.0049 and 0.0051

\({\text{\hat p}}\) is unknown (as we are interested in the sample size irrespective to the value of p ), then the sample size is (round up to the nearest integer!):

\({\text{n = }}\frac{{{\text{4}}{{\left[ {{{\text{z}}_{{\text{\alpha /2}}}}} \right]}^{\text{2}}}{\text{\hat p(1 - \hat p)}}}}{{{{\text{w}}^{\text{2}}}}}{\text{ - }}\frac{{{\text{4 \times 2}}{\text{.57}}{{\text{5}}^{\text{2}}}{\text{ \times 0}}{\text{.5(1 - 0}}{\text{.5)}}}}{{{\text{0}}{\text{.0}}{{\text{5}}^{\text{2}}}}}{\text{ = 2653}}\)

Note: If you use the critical value \({{\text{z}}_{{\text{\alpha /2}}}}{\text{ - 2}}{\text{.58}}\) , then you obtain \({\text{n - 2663}}\)

Hence the sample size is \({\text{n - 2653}}\)