A simple random sample of 100 8th graders at a large suburban middle school indicated that 81% 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.

The  interval is  

Step-by-step explanation:

From the question we are told that

      The  sample size is  

       The  population  proportion is

       The  confidence level is  C =  98%

The level of significance is mathematically evaluated as

    %

Here this level of significance represented the left and the right tail

The degree of  freedom is evaluated as

substituting value  

Since we require the critical value of one tail in order to evaluate the  98% confidence interval that estimates the proportion of them that are involved in an after school activity. we will divide the level of significance by 2

The  critical value of   and the evaluated degree of freedom is  

this is obtained from the critical value table  

The standard error is mathematically evaluated as

substituting value  

The 98%  confidence interval is evaluated as

substituting value  

c

step-by-step explanation:

c

10/60

step-by-step explanation:

if you divide 10 by 60 the answer is 0.166666 while if you divide or solve the other problems it equals 0.6