Question 9.7: A married couple tosses a coin after each dinner to determin......

(a) Using n = 98, we calculate an unbiased point estimate for p using \bar{x} =\hat{p }. Note that x_{i} = 1 if the wife has to wash the dishes.

\hat{p }=\frac{1}{n} \sum\limits_{i=1}^{n}{x_{i} }= \frac{1}{98} \cdot 59 =\frac{59}{98} \approx 0.602.

(b) Since n \hat{p }(1− \hat{p }) = 98 · 0.602 · 0.398 = 23.48 > 9 and p is sufficiently large, we can use the normal approximation to calculate a confidence interval for p. Using z_{{1−α}/{2}} = z_{0.975} = 1.96, we obtain

I_{l} (X) =0.602 − 1.96 \sqrt{\frac{0.602 · 0.398}{98} } = 0.505,

I_{u} (X) = 0.602 + 1.96 \sqrt{\frac{0.602 · 0.398}{98} } = 0.699.

This yields a confidence interval of [0.505, 0.699]. Note that the confidence interval does not contain p = 0.5 which is the probability that would be expected if the coin was fair. This is a clear sign that the coin might be unfair.

(c) If the coin is fair, we can use \hat{p } = 0.5 as our prior judgement. We would then need

n\geq \left[\frac{z_{{1−α}/{2}}}{\Delta } \right] ^{2} \hat{p} \left(1-\hat{p} \right)

\geq \left[\frac{1.96}{0.005 } \right] ^{2} 0.5 ^{2}= 38, 416

dinners to get the desired precision—which is not possible as this would constitute a time period of more than 100 years. This shows that the expectation of such a high precision is unrealistic and needs to be modified. However, as the results of (b) show, the coin may not be fair. If the coin is indeed unfair, we may have, for example, p = 0.6 and 1 − p = 0.4 which gives a smaller sample size. We can thus interpret the sample size as a conservative estimate of the highest number of dinners needed.