Question

The standard deviation of test scores on a certain achievement test is 11.5. A random sample of 90 scores on this test had a mean of 73.6. 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 answer 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,

= 73.6

s = 11.5

n = 90

Degrees of freedom = df = n - 1 = 90- 1 = 89

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,89 =1.662

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

= 1.662 * (11.5 / 90)

= 2.0

Margin of error = 2.0

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

- E < < + E

73.6 - 2.0 < < 73.6 + 2.0

71.6 < < 75.6

The lower limit =71.6

The upper limit =75.6