Answer to Question #314519 in Statistics and Probability for gideon

Question

 Television advertisers value the well-known Radio and TV awards scheme RTP as a measure of a TV/Radio show’s popularity among viewers. The RTP rating of a certain TV program is an estimate of the proportion of viewers, expressed as a percentage, who watch the program on a given day. In one survey, the RTP ratings team found that 101 out of 165 sampled families watched “The KSM Show” on the night of its premiere (the first time it is shown on television). In a separate survey several years later, the RTP ratings team found that 97 out of 180 sampled families watched “Music Music on TV3” on its premiere. (a) Estimate the true proportion p1 of all TV-viewing families who watched the premiere of “The KSM Show” and obtain a 95% confidence interval for p1. (b) Briefly explain what this 95% confidence interval means. (c) If the RTP team wanted to guarantee that all proportions are estimated to within ±0.05 with 95% confidence, how large should their samples be?

Answer 1

$a:\\\hat{p}_1=\frac{101}{165}=0.612121\\Confidence\,\,interval:\\\left( \hat{p}_1-z_{\frac{1+\gamma}{2}}\sqrt{\frac{\hat{p}_1\left( 1-\hat{p}_1 \right)}{n}},\hat{p}_1+z_{\frac{1+\gamma}{2}}\sqrt{\frac{\hat{p}_1\left( 1-\hat{p}_1 \right)}{n}} \right) =\\=\left( 0.612121-1.960\sqrt{\frac{0.612121\left( 1-0.612121 \right)}{165}},0.612121+1.960\sqrt{\frac{0.612121\left( 1-0.612121 \right)}{165}} \right) =\\=\left( 0.537771,0.686471 \right) \\b:\\With\,\,probability\,\,0.95 the\,\,true\,\,proportion\,\,lies\,\,within\,\,this\,\,interval\\c:\\z_{\frac{1+\gamma}{2}}\sqrt{\frac{\hat{p}_1\left( 1-\hat{p}_1 \right)}{n}}\leqslant 0.05\\\max \hat{p}_1\left( 1-\hat{p}_1 \right) =\frac{1}{2}\left( 1-\frac{1}{2} \right) =0.25\\1.960\sqrt{\frac{0.25}{n}}\leqslant 0.05\Rightarrow n\geqslant \frac{1.960^2}{0.05^2}\cdot 0.25=384.16\\n=385 is\,\,enough\\$