Families USA, a monthly magazine that discusses issues related to health and hea

Question

Families USA, a monthly magazine that discusses issues related to health and health
costs, surveyed 20 of its subscribers. It found that the annual health insurance premiums
for a family with coverage through an employer averaged $10,979. The standard deviation
of the sample was $1,000.
a. Based on this sample information, develop a 90 percent confidence interval for the
population mean yearly premium.
b. How large a sample is needed to find the population mean within $250 at 99 percent
confidence?

Explanation / Answer

a. The confidence interval is calculated as:

Mean ± t(/2) * st.dev. / n

where t(/2) is a number that comes from the t-distribution with n-1 = 20-1 = 19 degrees of freedom. In this case, t(0.10/2) is 1.73. The confidence interval is:

10,979 ± 1.73 * 1000/20

10,979 ± 387

So from $10,592 to $11,366.

b. We want to get the margin of error to ±250. The formula for the margin of error in a large sample is:

± z(/2) * st.dev. / n

So in this case, we want:

±250 = ± 2.58 * 1000 / n

±250 = ± 2,580 / n

Squaring both sides,

62,500 = 6,656,400 / n

n = 106.2 = 107

This is indeed a large enough number that we're okay with the large-sample formula.

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

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