Question

The following 5 questions are based on this information.

In a random sample of 225 individual tax returns in 2010,  72% (p bar=0.72) were filed electronically. The goal is to construct a 90% confidence interval of the proportion (p) of all individual tax returns that were filed electronically in 2010.

1) The standard error (SE) of p bar is

Select one:

a. 0.03

b. 0.72

c. 0.05

d. 0.20

2) The critical value (CV) needed for the 90% confidence interval estimation of p is

Select one:

a. 1.28

b. 0.05

c. 1.64

d. 0.1

3) The 90% confidence interval estimate of p  is

Select one:

a. 0.72 ± 0.05

b. 0.72 ± 0.1

c. 0.72 ± 0.15

d. 0.72 ± 0.03

4)Suppose around the period the above sample was taken, H & R Block (a tax preparation company ) claims that 69% of all individual tax returns are electronically filed. In light of the sample evidence and at the 10% level of significance,

Select one:

a. We cannot reject the commentator’s claim

b. We can reject the commentator’s claim

5) A tax accounting expert wishes to collect new random sample with the aim of building a new confidence interval at the 90% confidence level for p.

Using the current sample proportion (from the 225 individual tax returns) as a basis, what sample size (n) would the journalist require to achieve a 10% margin of error?

Select one:

a. 450

b. 78

c. 225

d. 55

Answer 1

Answer)

N = 225

P = 0.72

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 = 162

N*(1-p) = 63

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

1)

Standard error = √{p*(1-p)}/√n = 0.03

2)

Critical value z from z table for 90% confidence level is 1.64

3)

Margin of error (MOE) = Z*Standard error = 1.64*0.03 = 0.05

Option A

4)

Since the interval contains the claimed value 0.69

We cannot reject the claim

5)

MOE = z*√{p*(1-p)}/√n

0.1 = 1.64*√{0.72*0.28}/√n

N = 55