How common is SAT coaching? A random sample of students who took the SAT college entrance examination twice found that $427$of the respondents had paid for coaching courses and that the remaining $2733$ had not. ${}^{1+}$ Construct and interpret a $99\%$ confidence interval for the proportion of coaching among students who retake the SAT. Follow the four-step process.

x= Number of successes $=427$

n= Sample size $=427+2733=3160$

c=Confidence interval$=99\%=0.99$

p=Proportion of coaching among students who retake the SAT

Plan We are planning on calculation a one sample z-interval for a population proportion $p$.

Random condition: Satisfied, because the students come from a random sample.

$10\%$condition: Satisfied, because the sample of$3160$students are less than $10\%$of the entire population of students.

Large counts condition: Satisfied, because $n\hat{p}=$Number of successes $=427\ge 10$and $n(1-\hat{p})=$Number of failures $=2733$$\ge 10$

We note that all three conditions are satisfied.

localid="1650247130667" $$\begin{gathered} \hat{p}=\frac{x}{n} \\ =\frac{427}{427+2733} \\ =\frac{427}{3160} \\ \approx 0.1351 \end{gathered}$$

For confidence level$1-\alpha =99\%=0.99$determine $z_{\alpha /2}=z_{0.005}$using table A (look up $0.005$ in the table, the z-score is then the found$z$-score with opposite sign:

$z_{\alpha /2}=z_{0.005}=2.575$

The margin of error is then:

localid="1650247145286" $$\begin{gathered} E=z_{\alpha /2}·\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \\ =2.575·\sqrt{\frac{0.1351(1-0.1351)}{3160}} \\ \approx 0.0157 \end{gathered}$$

The confidence interval is then:

localid="1650247169639" $$\begin{gathered} 0.1194=0.1351-0.0157 \\ =\hat{p}-E<p<\hat{p}+E \\ =0.1351+0.0157 \\ =0.1508 \end{gathered}$$