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: [tex]\hat{p}=0.84[/tex]
Significance level : [tex]\alpha= 1-0.98=0.02[/tex]
Critical z-value : [tex]z_{\alpha/2}=2.33[/tex] (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 :-
[tex]\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
i.e. [tex]0.84\pm (2.33)\sqrt{\dfrac{0.84(1-0.84)}{100}}[/tex]
i.e. [tex]\approx0.84\pm 0.085[/tex]
i.e. [tex](0.84- 0.085,\ 0.84+ 0.085)=(0.755,\ 0.925)[/tex]
Hence, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity : a) (0.755, 0.925)
