Scores on an IQ test are normally distributed. A sample of 18 IQ scores had standard deviation s = 10. Construct a 95% confidence interval for the population variance σ2. Round the answers to two decimal places.

95% confidence interval for the population variance = [56.31 , 224.75].

Step-by-step explanation:

We are given that Scores on an IQ test are normally distributed. A sample of 18 IQ scores had standard deviation s = 10.

The pivotal quantity for 95% confidence interval for the population variance is given by;

              ~

95% confidence interval for the population variance is;

P(7.564 < < 30.19) = 0.95

P(7.564 < < 30.19) = 0.95

P( < < ) = 0.95

P( < < ) = 0.95

95% Confidence interval for = [ , ]

                                                 = [ , ]

                                                 = [ 56.31 , 224.75 ]

Therefore, 95% confidence interval for the population variance, is [ 56.31 , 224.75 ] .

1

x = 1

step-by-step explanation:

0 ÷ x = 0

0 ÷ 1 = 0

step-by-step explanation:

is their an equation anywhere i can take a   look at it and see if i am right.