Find $z_{\alpha/2}$ for an $84\%$ confidence interval.

$P(z>1.41)= 0.08$, so $z_{\alpha/2}=1.41$.Find $z_{\alpha/2}$ for a $93\%$ confidence interval.

$P(z>1.81)= 0.035$, so $z_{\alpha/2}=1.81$.How many times should we flip a coin to get a $98\%$ confidence interval for the probability that the coin lands heads with margin of error at most .01?

$n=(.5)(.5)\left({2.33\over .01}\right)^2=13572.25$ Round up to 13573.How large a sample would you need in order to estimate within $2\%$ the probability of rolling a "six" on a 6-sided die with $90\%$ confidence? In a pilot study, the die was rolled 200 times and landed on "six" 79 times.

$n=(79/200)(121/200)\left({1.645\over .02}\right)^2=1616.68$ Round up to 1617.In a sample of 100 M&M's, 8 of them were brown. Find a $94\%$ confidence interval for the proportion of M&M's that are brown.

$E=0.0510$ and $0.029 \lt p \lt0.131$. We are $94\%$ confident that the proportion of brown M&M's is between $2.9\%$ and $13.1\%$.The coin lands heads 48 times out of 100 tosses. Find a $97\%$ confidence interval for the probability that the coin lands heads.

$E=0.108$ and $0.372 \lt p \lt 0.588$. We are $97\%$ confident that the probability that the coin lands heads is in the interval $[0.372,0.588]$.The American Automobile Association claims that $54\%$ of fatal car/truck accidents are caused by driver error. A researcher studies 35 randomly selected accidents and finds that 14 were caused by driver error. Find a $92\%$ confidence interval for the proportion of fatal car/truck accidents caused by driver error.

$E=0.1449$ and $ 0.255 \lt p \lt 0.545$. We are $92\%$ confident that the proportion of fatal car/truck accidents caused by driver error is between 0.255 and 0.545.A survey showed that among 785 randomly selected subjects who completed four years of college, $18.3\%$ smoke. Find a $96\%$ confidence interval for the proportion of those with four years of college who smoke.

$E=0.028$ and $0.155 \lt p \lt 0.211$ We are $96\%$ confident that the probability that the proportion of those with four years of college who smoke is in the interval $[15.5\%,21.1\%]$.In a random sample of 250 children, it was found that 87 children were dressed as movie characters for Halloween. Find a $92\%$ confidence interval for the proportion of children who dress as movie characters for Halloween.

$E=0.0527$ and $0.295 \lt p \lt 0.401$ We are $92\%$ confident that the proportion of children who dress as movie characters for Halloween is between 0.295 and 0.401.A $90\%$ confidence interval for the proportion of the population in favor of an issue is $.41 \lt p \lt .47$. Give the point estimate and the maximum error of the estimate.

$\widehat{p} = (.41+.47)/2=.44$ and $E=(.47-.41)/2=.03$For a random sample of students, a $95\%$ confidence interval for the proportion who prefer chocolate ice cream is $[0.31, 0.39]$. Find the point estimate and the margin of error.

$\widehat{p} = (.31+.39)/2=.35$ and $E=(.39-.31)/2=.04$Beth and Lori share a digital music player that has a feature that randomly selects which song to play. A total of 2345 songs were loaded onto the player, some by Beth and the rest by Lori. Suppose that when the player was in the random-selection mode, 36 of the first 50 songs selected were songs loaded by Beth.

Construct and interpret a $90\%$ confidence interval for the proportion of songs on the player that were loaded by Beth.

How large a sample would be needed to estimate the proportion of songs loaded by Beth with $99\%$ confidence and margin of error at most $5\%$?

- $E=0.1044$ and $0.616 \lt p \lt 0.824$. We are $90\%$ confident that the proportion of songs on the player that were loaded by Beth is between $61.6\%$ and $82.4\%$.
- $n=(.72)(.28)\left({2.58\over .05}\right)^2=536.8$ Round up to 537.