Answer to Question #314156 in Statistics and Probability for Bless

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) Let p2 be the true proportion of all TV-viewing families who watched the premiere of “Music Music on TV3”. Estimate p1 − p2 and obtain a 95% confidence interval for p1 − p2.

(b) From the results in part (a), can you tell whether the premiere of “The KSM Show” was more popular that the premiere of “Music Music on TV3”? Give a brief justification of your answer.

Answer 1

$a :\\\hat{p}_1=\frac{101}{165}=0.612121\\\hat{p}_2=\frac{97}{180}=0.538889\\\hat{p}=\frac{101+97}{165+180}=0.573913\\Confidence\,\,interval:\\\left( \hat{p}_1-\hat{p}_2-\sqrt{\hat{p}\left( 1-\hat{p} \right) \left( \frac{1}{n_1}+\frac{1}{n_2} \right)}z_{\frac{1+\gamma}{2}},\hat{p}_1-\hat{p}_2+\sqrt{\hat{p}\left( 1-\hat{p} \right) \left( \frac{1}{n_1}+\frac{1}{n_2} \right)}z_{\frac{1+\gamma}{2}} \right) =\\=\left( 0.612121-0.538889-\sqrt{0.573913\left( \frac{1}{165}+\frac{1}{180} \right)}\cdot 1.9600,0.612121-0.538889+\sqrt{0.573913\left( \frac{1}{165}+\frac{1}{180} \right)}\cdot 1.9600 \right) =\\=\left( -0.0868013,0.233265 \right)$

b: Since the confidence interval contains 0, we cannot conclude that one proportion is greater than another.