91影视

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 $=427鈮�10$and $n(1-\hat{p})=$Number of failures $=2733$$鈮�10$

We note that all three conditions are satisfied.

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

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

$z_{伪/2}=z_{0.005}=2.575$

The margin of error is then:

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

The confidence interval is then:

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