Question

A Simple random sample of 100 8th graders at a large suburban middle school indicated that 84% of them are involved with some type of after school activity. Find the 98% confidence interval that estimates the proportion of them that are involved in an after school activity.
a) (0.755, 0.925)
b) (0.755, 0.725)
c) (0.805, 0.810)
d) (0.675, 0.925)
e) (0.655, 0.875)
f) None of the above

Answer 1

Answer: a) (0.755, 0.925)

Step-by-step explanation:

Let p be the population proportion of 8th graders are involved with some type of after school activity.

As per given , we have

n= 100

sample proportion: $\hat{p}=0.84$

Significance level : $\alpha= 1-0.98=0.02$

Critical z-value : $z_{\alpha/2}=2.33$  (using z-value table)

Then, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity will be :-

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

i.e. $0.84\pm (2.33)\sqrt{\dfrac{0.84(1-0.84)}{100}}$

i.e. $\approx0.84\pm 0.085$

i.e. $(0.84- 0.085,\ 0.84+ 0.085)=(0.755,\ 0.925)$

Hence, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity : a) (0.755, 0.925)