## Related questions with answers

A one-sided confidence interval for p can be expressed as $p<\hat{p}+E \text { or } p>\hat{p}-E$, where the margin of error E is modified by replacing $z_{\alpha / 2}$ with $z_{\alpha}$. If Air America wants to report an on-time performance of at least x percent with 95% confidence, construct the appropriate one-sided confidence interval and then find the percent in question. Assume that a simple random sample of 750 flights results in 630 that are on time.

Solution

VerifiedThe best - point estimate of $p$ is the number of successes $x$ divided by the sample size $n$:

$\hat{p}=\dfrac{x}{n}=\dfrac{630}{750}\approx 0.84$

Determine $z_{\alpha/2}$ using table A-2:

$z_{\alpha/2}=z_{0.05}=1.645$

The margin of error is then:

$E=z_{\alpha/2}\cdot \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}=1.645\cdot \sqrt{\dfrac{0.84(1-0.84)}{750}}\approx 0.0220$

The confidence interval then becomes:

$81.80\%=0.8180=0.84-0.0220=\hat{p}-E<p$

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