Grade 70 80 60 90 75

n = 5

xbar = 75

s = 11.18

a. Determine a point estimate for the variance of the population.

s2= 11.182 ?

**125**

b. Determine a 95% confidence interval for the variance of the population.

(n-1)s^2/ x^2[sub:12t42v3o]?/2[/sub:12t42v3o] ? ?2 ? (n-1)s^2/x^2[sub:12t42v3o](1-?/2)[/sub:12t42v3o] n-1 = 5-1 =4 df and ? = .05

x^2[sub:12t42v3o].975[/sub:12t42v3o]= .484 ? (n-1)s^2/?2 ? 11.143 = x^2[sub:12t42v3o].025[/sub:12t42v3o]

(5-1)125/ 11.143 ? ?2 ? (5-1)125/.484 =

**44.87 ? ?2 ? 1033.06**

c. (This is where I'm lost. Am I on the right track?) At 90% confidence, test to determine if the variance of the population is significantly

more than 50.

H0 : ?2 ? .10 df = 5-1 = 4 x2.10 = 7.779

H? : ?2 > .10

Reject if x2 ? 50