Question 9.9: To estimate the audience rate for several TV stations, 3000 ......

(a) Whether the i th household has switched on the TV and watches “Germany’s next top model” (GNTM) relates to a random variable X_{i} with

X_{i} = 1 : if TV switched on and household watching GNTM
X_{i} = 0 : if TV switched off or household watches another show.

It follows that X = \Sigma ^{2500}_{i=1} X_{i} is the random variable describing the number of TVs, out of 2500 TVs, which are switched on and show GNTM. Since the X_{i} ’s can be assumed to be i.i.d., we can say that X follows a binomial distribution, i.e. X ∼ B(2500; p) with p unknown. The length of the confidence interval for p,

\left[\hat{p} – z_{{1−α}/{2}}\sqrt{\frac{\hat{p}\left(1-\hat{p}\right) }{n} }, \hat{p}+ z_{{1−α}/{2}}\sqrt{\frac{\hat{p}\left(1-\hat{p}\right) }{n} }\right],

is

L =2 z_{{1−α}/{2}}\sqrt{\frac{\hat{p}\left(1-\hat{p}\right) }{n} }.

Unfortunately, \hat{p}\left(1-\hat{p}\right) is unknown but \hat{p}\left(1-\hat{p}\right) ≤ \frac{1}{4} because the maximum value for \hat{p}\left(1-\hat{p}\right) is 0.25 if \hat{p} = 0.5. Hence

L ≤ 2 z_{{1−α}/{2}}\sqrt{\frac{\frac{1}{4} }{n}} = \frac{1.96}{\sqrt{2500} }= 0.0392.

This means the precision is half the length, i.e.±0.0196 = ±1.96 %, in the worst case.

(b) There is the danger of selection bias. A total of 500 households refused to take part in the study. It may be that their preferences regarding TV shows are different from the other 2500 households. For example, it may well be possible that those watching almost no TV refuse to be included; or that those watching TV shows which are considered embarrassing by society are refusing as well. In general, missing data may cause point estimates to be biased, depending on the underlying mechanism causing the absence.