Families USA, a monthly magazine that discusses issues related to health and hea
ID: 3381276 • Letter: F
Question
Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 22 of its subscribers. It found that the annual health insurance premiums for a family with coverage through an employer averaged $10,800. The standard deviation of the sample was $1,060. (Use z Distribution Table.)
a. Based on this sample information, develop a 90% confidence interval for the population mean yearly premium. (Round your answers to the nearest 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 $240 at 90% confidence? (Round up your answer to the next whole number.)
Sample size
Explanation / Answer
a)
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.05
X = sample mean = 10800
z(alpha/2) = critical z for the confidence interval = 1.64
s = sample standard deviation = 1060
n = sample size = 22
Thus,
Margin of Error E = 370.6281253
Lower bound = 10429.37187
Upper bound = 11170.62813
Thus, the confidence interval is
(10429 , 11171) [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) = 1.64
Also,
s = sample standard deviation = 1060
E = margin of error = 240
Thus,
n = 52.46587778
Rounding up,
n = 53 [ANSWER]
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