Asimple 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

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:

Significance level :

Critical z-value :  (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 :-

i.e.

i.e.

i.e.

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

answer: 1,380

step-by-step explanation:

1: you turn %15 into a decimal by moving the decimal point two places to the left.

2: and then you multiply .15 by 1,200 , which will get you 180.00

3: from there you will add 180 to 1,200

4: which will get you 1,380

hope it *im kind of slow so idk if my answer is right*

n

step-by-step explanation: