Question

Exhibit 11-1 Last year, the standard deviation of the ages of the students at UA was 1.81 years. Recently, a sample of 10 students had a standard deviation of 2.1 years. We are interested in testing to see if there has been a significant change in the standard deviation of the ages of the students at UA. Refer to Exhibit 11-1. The test statistic is

Answer 1

Answer:

test statistic = 12.115

Step-by-step explanation:

Given data :

std = 2.1 years

n = 10

( std )^2 = 4.41

determine the test statistic

apply a two-tailed test ( chi squared test for one population variance )

test statistic

λ^2 = $\frac{(n-1)s^2}{3.2761}$   ( Referring to Exhibit 11-1 )

      = ( 10 - 1 )*(2.1)^2 / 3.2761

      = 12.115