Question

A study was done by researchers at a university to determine the percentage of all college students that exercise at least once per week. The study found that 411 students reported exercising at least once per week out of 800 surveyed. Based on this sample, construct a 96% confidence interval for the proportion of all college students that exercise "regularly" . Show formula, substitution, and result

Answer 1

Solution :

Given that,

n = 800

x = 411

$\hat p$ = x / n = 411/ 800 = 0.514

1 -$\hat p$ = 1 - 0.514 = 0.486

At 96% confidence level the z is ,

$\alpha$ = 1 - 96% = 1 - 0.96 = 0.04

$\alpha$ / 2 = 0.04 / 2 = 0.02

Z_$\alpha$/2 = Z_0.02 = 2.054

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$(($\hat p$ * (1 - $\hat p$ )) / n)

= 2.054 * ($\sqrt$((0.514 * 0.486) / 800) = 0.036

A 96 % confidence interval for proportion p is ,

$\hat p$ - E < P <$\hat p$ + E

0.514 - 0.036 < p < 0.514 + 0.036

0.478< p < 0.550

(0.478,0.550)

96% confidence interval for the proportion of all college students that exercise "regularly" (0.478,0.550)