Question

The high price of medicines is a source of major expense for those seniors in the United States who have to pay for these medicines themselves. A random sample of 2000 seniors who pay for their medicines showed that they spent an average of $4100 last year on medicines with a standard deviation of $600. Make a 99% confidence interval for the corresponding population mean

Answer 1

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

From given data, we have

Xbar = 4100

S = 600

n = 2000

df = n – 1 = 1999

Confidence level = 99%

Critical t value = 2.5783

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 4100 ± 2.5783*600/sqrt(2000)

Confidence interval = 4100 ± 34.5914

Lower limit = 4100 - 34.5914 = 4065.41

Upper limit = 4100 + 34.5914 =4134.59

Confidence interval = ($4,065.41, $4,134.59)