Question

The standard deviation of test scores on a certain achievement test is 11.1. A random sample of 60 scores on this test had a mean of 75.4. Based on this sample, find a 95% 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. What is the lower limit of the 95% confidence interval? What is the upper limit of the 95% confidence interval? х ?

Answer 1

Solution :

Given that,

Z$\alpha$/2 = Z_0.025 = 1.96

Margin of error = E = Z$\alpha$/2 * ( $\sigma$ /$\sqrt$n)

= 1.96 * ( 11.1 /  $\sqrt$60 )

= 2.8

At 95% confidence interval estimate of the population mean is,

  ± E

T

= 75.4   ± 2.8

= ( 72.6, 78.2 )

lower limit = 72.6

upper limit = 78.2