Question

In a random survey of 300 self-employed individuals, 18 report having had their tax returns audited by the IRS in the past year.

Does your confidence interval provide evidence that at least 5% of self-employed
individuals had their tax returns audited by the IRS last year? Yes No
Explain your answer.

Answer 1

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

x = 18

n = 300

P = x/n = 18/300 = 0.06

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.06 ± 1.96* sqrt(0.06*(1 – 0.06)/300)

Confidence Interval = 0.06 ± 1.96* 0.0137

Confidence Interval = 0.06 ± 0.0269

Lower limit = 0.06 - 0.0269 = 0.0331

Upper limit = 0.06 + 0.0269 = 0.0869

Confidence interval = (0.0331, 0.0869)

Does your confidence interval provide evidence that at least 5% of self-employed individuals had their tax returns audited by the IRS last year?

Answer: No

..because above confidence interval includes values less than 5% or 0.05.