Coaching and SAT scores (10.1) What proportion of students who take the SAT twice are coached? To answer this question, Jannie decides to construct a $99\%$confidence interval. Her work is shown below. Explain what’s wrong with Jannie’s method.

A $99\%$CI for $p_1-p_2$is

$(0.135-0.865)\pm 2.575\sqrt{\frac{0.135(0.865)}{3160}+\frac{0.865(0.135)}{2733}}$$=-0.73\pm 0.022=(-0.752,-0.708)$

We are $99\%$ confident that the proportion of students taking the SAT twice who are coached is between $71$ and $75$ percentage points lower than students who aren’t coached.

Summary statistic is

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

$\hat{p}\pm z_{\alpha /2}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

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:

$CI=\hat{p}\pm z_{\alpha /2}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$=0.135\pm 2.576\times \sqrt{\frac{0.135(1-0.135)}{3160}}$

$=(0.119,0.151)$

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 $0.151$.