You play a game and win 136 out of 270 times.

Estimate the probability of winning the game.

$\widehat{p} = 136\,/\,270 \doteq 0.5037$Find a 95% confidence interval for the probability of winning the game.

We first check the requirement that $np \ge 5$ and $nq \ge 5$ by estimating $p$ with the sample proportion $\widehat{p}$ and ensuring that we see at least $5$ successess and $5$ failures. Of course, here, we see $136$ successes (wins) and $134$ failures (losses), so the requirements are met.

Now note, for 95% confidence we have $z_{\alpha/2} = 1.96$, so the confidence interval is given by:

$$\frac{136}{270} \pm 1.96 \sqrt{\frac{\left(\frac{136}{270}\right) \left(\frac{134}{270}\right)}{270}}$$ or approximately $$(0.4441, 0.5633)$$

Find $z_{\alpha/2}$ for a 90% confidence interval for a proportion.

$$z_{\alpha/2} = 1.645$$ Note, on a TI-83 calculator,`invNorm((1 - 0.90)/2) = -1.645`

A CBS News/New York Times poll found that 329 out of 763 adults said they would travel to outer space in their lifetime, given the chance. Estimate the true proportion of adults who would like to travel to outer space with 92% confidence.

Check requirements: Substituting $\widehat{p}$ for $p$, we have $np \approx 329 \ge 5$ and $nq \approx 434 \ge 5$, so the requirements are met.

$\widehat{p} = 329\,/\,763$

$n = 763$

$92\% \rightarrow z_{\alpha/2} = 1.751$,

Thus, the confidence interval is given by:

$$\frac{329}{763} \pm 1.751 \sqrt{ \frac{ \left(\frac{329}{763}\right) \left(\frac{434}{763}\right)}{763}}$$

$$= (0.3998,0.4626)$$The Gallup Poll found that 27% of adults surveyed nationwide said they had personally been in a tornado. How many adults should be surveyed to estimate the true proportion of adults who have been in a tornado with a 95% confidence interval 5% wide?

Recall $ME = z_{\alpha/2} \sqrt{\frac{\widehat{p}\widehat{q}}{n}}$, so solving for $n$ we have $$n = \widehat{p}\widehat{q} \left( \frac{z_{\alpha/2}}{ME} \right)^2$$We don't know $\widehat{p}$ or $\widehat{q}$, as the survey hasn't happened yet -- but we can approximate these with the Gallup results. Thus,

$$n = (0.27)(0.73) \left(\frac{1.960}{0.05}\right)^2 \doteq 302.87$$So one should survey 303 adults. (Actually, one should probably survey more than this to take into account response bias.)

A recent study indicated that 29% of the 100 women over age 55 in the study were widows.

Find a 90% confidence interval for the true proportion of women over age 55 who are widows.

Checking requirements, note that $np = (100)(0.29) = 29 \ge 5$ and $nq = (100)(0.71) = 71 \ge 5$, so the requirements are met.

$90\%$ confidence $\rightarrow z_{\alpha/2} = 1.645$, so the confidence interval is given by

$$0.29 \pm 1.645 \sqrt{\frac{(0.29)(0.71)}{100}}$$ $$= (0.2154, 0.3646)$$How large a sample must one take to be 90% confident that the estimate is within 0.05 of the true proportion of women over age 55 who are widows?

$$n = pq\left(\frac{z_{\alpha/2}}{ME}\right)^2$$ Using the recent study percentages as approximations for $p$ and $q$, we have $$n = (0.29)(0.71)\left(\frac{1.645}{0.05}\right)^2 \doteq 222.86$$ Thus, one should take a sample of size $223$ women over $55$.If no estimate of the sample proportion is available, how large should the sample be to be 90% confident that the estimate is within 0.05 of the true proportion?

In the absence of an estimate of the sample proportion, we error on the side of being overly conservative. The maximum product of $pq$ is $1/4$, so take

$$n = \frac{1}{4} \left(\frac{z_{\alpha/2}}{ME}\right)^2 = 270.6$$Thus, one should sample 271 women over 55.