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.

An organization advertises that on a given poll, 43% answered "yes" to the question "Would you rather have a boring job than no job?", with a margin of error of $\pm1\%$. What did the organization fail to reveal?

The confidence level for the margin of error reported.Upon taking a sample to estimate a population proportion, why is it better to report a confidence interval than $\widehat{p}$, the best point-estimate for this proportion.?

Just reporting $\widehat{p}$ does not indicate the accuracy of the estimate.Find the indicated $z$-scores:

- Find the $z$-scores associated with $92\%$ confidence
- Find the $z$-scores associated with $99.5\%$ confidence

- $\pm1.7507$
- $\pm2.8070$

Express the confidence interval $0.200 \lt p \lt 0.500$ in the form of $\widehat{p} \pm E$

$0.350 \pm 0.150$Find the confidence interval for a proportion if $\widehat{p} = 0.222$ and the margin of error is $0.044$.

$(0.178,0.266)$If a sample is used to estimate a population proportion $p$, find the margin of error $E$ that corresponds to $n = 1000, x = 400$, and $95\%$ confidence.

$0.0304$Construct a $95\%$ confidence interval to estimate a population proportion $p$ if $n=200, x=40$.

$0.145 \lt p \lt 0.255$Construct a $99\%$ confidence interval for a population proportion $p$ if $n=1236, x = 109$

$(0.0674,0.109)$What sample size should be used to estimate a population proportion within $0.045$ with $95\%$ confidence?

$475$What sample size should be used to estimate a population proportion within $2\%$, with $99\%$ confidence, when a prior study estimated $\widehat{p} = 0.14$?

$1998$A medication is suspected of increasing the likelihood of conceiving a girl. Of $574$ pregnancies where the mother was taking this medication during her pregnancy, $525$ of them gave birth to a girl. Construct a $95\%$ confidence interval for the proportion of births that result in a girl when the mother is taking this medication.

$0.892 \lt p \lt 0.937$In one of Mendel's famous genetics experiments with peas, he predicted that $25\%$ of offspring peas would be yellow. He instead saw 152 yellow peas and 428 green peas. Find a $95\%$ confidence interval estimate for the percentage of yellow peas. Do the results contradict his hypothesis?

$0.226 \lt p \lt 0.298$; These results do not contradict his hypothesis, as $0.25$ is included in the confidence interval.In a study of $420,095$ cell phone users, $135$ developed brain cancer or cancer of the nervous system. Prior to this study, it was found that the rate of such cancers was $0.034\%$ for people not using cell phones. Construct a $95\%$ confidence interval for the proportion of cell phone users that develop such cancers. Is there a significant difference between cell phone users and people that don't use cell phones?

$0.000267 \lt p \lt 0.000376$; There does not appear to be a significant difference as $0.00034$ is included in the confidence interval.Assuming we wish to estimate the proportion of American adults who use the internet within $2\%$ and be $99\%$ confident in our results, how many randomly selected adults should we survey?

$4147$