Question

The standard deviation of test scores on a certain achievement test is 10.9. A random sample of 60 scores on this test had a mean of 72.5. Based on this sample, find a 90% confidence interval for the true mean of all scores. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)

What is the lower limit of the 90% confidence interval? What is the upper limit of the 90% confidence interval?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 72.5

sample standard deviation = s = 10.9

sample size = n = 60

Degrees of freedom = df = n - 1 =60-1=59 90% confidence level the t is ,

= 1 - 90% = 1 - 0.9 = 0.1

/ 2 = 0.1/ 2 = 0.05

t /2,df = t0.05,59 = 1.671

Margin of error = E = t/2,df * (s /n)

= 1.671 * (10.9 / 60)

E = 2.351

The 90% confidence interval estimate of the population mean is,

- E < < + E

72.5 - 2.351 < < 72.5 + 2.351

70.1 < < 74.9

(70.1,74.9)

lower limit of the 90% confidence interval is 70.1

upper limit of the 90% confidence interval is 74.9