Question

My grandson plays soccer and he had a game a couple of weeks ago. The age group is 6 to 9 years of age and in this group there are a total of 12 girls and boys. Of the 12 soccer players there are 4 kids who are new to the soccer field. The opposing team they played were ages 7 to 10 years of age and this team has only 10 players, 6 of which are also new to the field. Propose a confidence level at 75% for the population proportion of soccer members who have never played soccer before.

Answer 1

Answer)

Total = 12 + 10 = 22

N = 22

Of these 4 + 6 are new that is they have never played soccer before

So, P = 10/22

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 10

N*(1-p) = 12

Both the conditions are met so we can use standard normal z table to estimate the interval

Critical value z from z table for 75% confidence level is 1.15

Margin of error (MOE) = Z*√{P*(1-P)}/√N

MOE =  0.12208278897

Interval is given by

(P-MOE, P+MOE)

(0.33246266557, 0.57662824351)