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Summary statistic is

The formula to construct the confidence interval for one-sample proportion is:

$\hat{p}卤z_{伪/2}脳\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Here, it is required to compute the confidence interval for one-sample proportion instead of difference in proportions. The mistake done by Jannie is that she has computed confidence interval for difference in proportions instead of one-sample proportion.

The calculation for the required confidence interval could be done as:

The sample proportion is calculated as:

$\hat{p}=\frac{x}{n}$

$=\frac{427}{427+2733}$

$=0.135$

The z-score at $99\%$confidence level is $2.576$.

The confidence interval is:

$CI=\hat{p}卤z_{伪/2}脳\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$=0.135卤2.576脳\sqrt{\frac{0.135(1-0.135)}{3160}}$

$=(0.119,0.151)$

Thus, the required confidence interval is $(0.119,0.151)$.

Interpretation:

There is $99\%$probability that the proportion of student that are taking the SAT twice are being coaches lies between $0.119$and $0.151$.