A one-sided confidence interval for p can be written as $p \hat { p } - \mathrm { E }$ where the margin of error E is modified by replacing $z _ { \alpha / 2 }$ with $z _ { \alpha }$ If a teacher wants to report that the fail rate on a test is at most x with 90% confidence, construct the appropriate one -sided confidence interval. Assume that a simple random sample of 74 students results in 8 who fail the test. A) p

Question

A one-sided confidence interval for p can be written as $p < \hat { p } + \mathrm { E } \text { or } p > \hat { p } - \mathrm { E }$ where the margin of error E is modified by replacing $z _ { \alpha / 2 }$ with $z _ { \alpha }$ If a teacher wants to report that the fail rate on a test is at most x with 90% confidence, construct the appropriate one -sided confidence interval. Assume that a simple random sample of 74 students results in 8 who fail the test. A) p<0.062 B) p<0.154 C) p>0.062 D) p<0.167

Quizplus · Accepted Answer

To construct a one-sided confidence interval for the fail rate $p$ with 90% confidence, we first calculate the sample proportion $\hat{p} = \frac{8}{74}$. Then, we find $z_{\alpha}$ for a 90% confidence level, which is $z_{0.10}$ (since it's one-sided, we don't divide $\alpha$ by 2). The value of $z_{0.10}$ is approximately 1.28. The margin of error $E$ is calculated using the formula $E = z_{\alpha} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $n$ is the sample size. Plugging in the values, we get $E \approx 1.28 \sqrt{\frac{(8/74)(66/74)}{74}}$, which gives a margin of error that, when added to $\hat{p}$, results in a value less than 0.154. Therefore, the correct one-sided confidence interval to report that the fail rate is at most $x$ with 90% confidence is $p < 0.154$.

A One-Sided Confidence Interval for P Can Be Written As p