Families USA, a monthly magazine that discusses issues related to health and hea
ID: 3134779 • Letter: F
Question
Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 21 of its subscribers. It found that the annual health insurance premiums for a family with coverage through an employer averaged $10,900. The standard deviation of the sample was $1,020.
(a) Based on this sample information, develop a 98% confidence interval for the population mean yearly premium. (Round up your answers to the next whole number.) Confidence interval for the population mean yearly premium is between $ and $ .
(b) How large a sample is needed to find the population mean within $230 at 90% confidence? (Round up your answer to the next whole number.) Sample size:
Explanation / Answer
a)
Note that
Margin of Error E = t(alpha/2) * s / sqrt(n)
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.01
X = sample mean = 10900
t(alpha/2) = critical t for the confidence interval = 2.527977003
s = sample standard deviation = 1020
n = sample size = 21
df = n - 1 = 20
Thus,
Margin of Error E = 562.6828043
Lower bound = 10337.3172
Upper bound = 11462.6828
Thus, the confidence interval is
( 10337.3172 , 11462.6828 ) [ANSWER]
********************
b)
Note that
n = z(alpha/2)^2 s^2 / E^2
where
alpha/2 = (1 - confidence level)/2 = 0.05
Using a table/technology,
z(alpha/2) = 1.644853627
Also,
s = sample standard deviation = 1020
E = margin of error = 230
Thus,
n = 53.21072608
Rounding up,
n = 54 [ANSWER]
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