Families USA , a monthly magazine that discusses issues related to health and he
ID: 3381647 • Letter: F
Question
Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 19 of its subscribers. It found that the annual health insurance premiums for a family with coverage through an employer averaged $10,500. The standard deviation of the sample was $1,060. (Use z Distribution Table.)
Based on this sample information, develop a 90% confidence interval for the population mean yearly premium. (Round your answers to the nearest whole number.)
How large a sample is needed to find the population mean within $250 at 99% confidence? (Round up your answer to the next whole number.)
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.)
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 = 10500
z(alpha/2) = critical z for the confidence interval = 1.64
s = sample standard deviation = 1060
n = sample size = 19
Thus,
Margin of Error E = 398.8163118
Lower bound = 10101.18369
Upper bound = 10898.81631
Thus, the confidence interval is
( 10101.18369 , 10898.81631 ) [ANSWER]
******************
B)
Note that
n = z(alpha/2)^2 s^2 / E^2
where
alpha/2 = (1 - confidence level)/2 = 0.005
Using a table/technology,
z(alpha/2) = 2.58
Also,
s = sample standard deviation = 1060
E = margin of error = 250
Thus,
n = 119.6660966
Rounding up,
n = 120 [ANSWER]
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