Answer to Question #183463 in Statistics and Probability for Amoah Henry

Question

 A marketing company, Groupe RM, has developed an index that seeks to determine the audience size and composition of television programming using a rating system. In one survey, Groupe RM found that 101 out of 165 sampled people watched “Sinking Sands” on the night of its premiere. In a separate survey later in the year, Groupe RM found that 97 out of 180 sampled people watched “Adams Apple” by Sparrows Production on its premiere.

(a) Estimate the true proportion p1 of all TV-viewing people who watched the premiere of Sinking Sands and obtain a 95% confidence interval for p1.

(b) Briefly explain what this 95% confidence interval means.

(c) If the Groupe RM company wanted to guarantee that all proportions are estimated to within ±0.05 with 95% confidence, how large should their samples be?

(d) Let p2 be the true proportion of all TV-viewing families who watched the premiere of Adams Apple. Estimate p1 −p2 and obtain a 95% confidence interval for p1 −p2.

Answer 1

a.) Proportion $p_1 = \dfrac{101}{165}$

95% confidence interval for $p_1$ can be calculated as $p_1 \pm z\sqrt{\dfrac{p_1(1-p_1)}{n}}$

= $0.61 \pm 1.96\sqrt{\dfrac{0.61 \times 0.38}{165}}$

$= 0.61 \pm 0.07$

b.)  A confidence interval displays the probability that a parameter will fall between a pair of values around the mean. Confidence intervals measure the degree of uncertainty or certainty in a sampling method.

c.) Sample size can be calculated as $n = p_1(1-p_1)(\dfrac{z}{E})^2$

$= 0.61 \times 0.38 \times (\dfrac{1.96}{0.05})^2$

$= 356.19$

d.) Proportion $p_2$ can be calculated as $\dfrac{97}{180}$ $= 0.53$

$p_1 - p_2 = 0.61- 0.53 = 0.08$

95% confidence interval for $p_1 - p_2 = (p_1-p_2) \pm z\sqrt{\dfrac{p_1(1-p_1)}{n}+{\dfrac{p_2(1-p_2)}{n}}}$

$= 0.08 \pm 1.96 \sqrt{\dfrac{0.61\times 0.39}{165}+\dfrac{0.53 \times 0.47}{180}}$

$= 0.08 \pm 2.01$