Question

In a sample of 300 students, 68% said they own an iPod and a smart phone. Compute a 97% confidence interval for the true percent of students who own an iPod and a smartphone.

Answer 1

$p'=0.68$

$q'=1-p'=1-0.68=0.32$

Since CL = 0.97

$\alpha=1-0.97=0.03$

$\frac{\alpha}{2}=\frac{0.03}{2}=0.015$

The area to the left of Z is 0.015 and the area to the right of Z is 1 – 0.015 = 0.985.

$Z=2.17$

$CI=p'\pm Z_\frac{\alpha}{2}\sqrt{\frac{p'q'}{n}}$

$CI=0.68\pm 2.17\sqrt{\frac{0.68\times0.32}{300}}$

$CI=0.68\pm 0.0269$

$CI=(0.68-0.0269), (0.68+0.0269)$

$CI=0.6531, 0.7069$

This means 97% confidence interval for the true percent of students who own an iPod and a smartphone is between 0.6531 and 0.7069.