The grades of a sample of 5 students, selected from a large population, are given below.
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
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