Question

1. Exhibit 4
   A 2016 study of 70 elementary school kids found that 30 of them spent significant time outside during the summer. A similar study in 1980 of 70 kids had found that 40 of them spent significant time outside during the summer.

   What is the point estimate of difference in proportion of kids who played outside in 1980 and 2016?

   		0.10
   		0.14
   		0.18
   		0.25

2 points   

QUESTION 15

1. Refer to Exhibit 4 in question 14 to answer this question.

   What is 95% confidence interval estimate of difference in proportion of kids who played outside in 1980 and 2016 (to 3 decimal places)?

   		-0.012 to 0.282
   		0.020 to 0.260
   		0.051 to 0.229
   		-0.095 to 0.375

Answer 1

Answer)

P1 = 40/70

P2 = 30/70

N1 = N2 = 70

-

Point estimate = (40/70) - (30/70) = 0.14

-

First we need to check the conditions of normality that is if n1*p1 and n2*p2 both are greater than 5 or not

N1*P1 = 40

N2*P2 = 30

BOTH are greater than 5

Conditions are met so we can use standard normal z table to estimate the interval.

Critical value z for a 95% confidence level, from z table is = 1.96

Margin of error (MOE) = 1.96*Standard error

Standard error = √{p1*(1-p1)}/√n1 + √{p2*(1-p2)}/n2}

MOE = 0.16395121220

Confidence interval is given by

Mean - MOE to Mean + MOE

-0.021 to 0.307

Closest option is option first