Question

There are 350 families in one area in Brooks city. A poll of 50 families reveals the mean annual charity contribution is $550 with a standard deviation of $85.

The same study of charity contributions revealed that 20 of the 50 families sampled do charity regularly.

1. Should the finite-population correction factor be used? Why or why not?
2. Construct the 95 percent confidence interval for the proportion of families doing charity regularly.

Answer 1

1)

Yes as the finite population is given

Finite-population correction factor is given by

√{(N-n)/(N-1)}

N = population size = 350

n = sample size = 50

2)

N = 50

P = 20/50

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 = 20

N*(1-p) = 30

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

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

Margin of error(MOE) = Z*STANDARD ERROR

Standard error =( √P*(1-p)/√n )*√{(N-n)/(N-1)}

N = population size = 350

n = sample size = 50

MOE = 0.12589967352

Confidence interval is given by

(P-MOE, P+MOE)

(0.27410032647, 0.52589967352)