A consumer group wants to estimate the mean electric bill for the month of July

Question

A consumer group wants to estimate the mean electric bill for the month of July for single-family homes in a large city. Based on studies conducted for other cities, the standard deviation is assumed to be $18. The group wants to estimate, with 99% confidence, the mean bill for July to within plusminus $3. a. What sample size is needed? b. If 95% confidence is desired, how many homes need to be selected? a. The sample size required for 99% confidence is (Round up to the nearest integer.) b. The sample size required for 95% confidence is (Round up to the nearest integer.)

Explanation / Answer

standard deviation = 18

margin of error = ± 3

let sample size be n

standard deviation of mean = 18/n

a)

margin of error for 99% confidence interval is 2.575*(18/n)

2.575*(18/n) = 3 =>n = 2.575*18/3 =15.45

n = 238.7 ~239

sample size needed for 99% connfidence is 239

b)

margin of error for 95% confidence interval is 1.96*(18/n)

1.96*(18/n) = 3 =>n =1.96*18/3 =11.76

n = 138.29 ~139

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.