# statistics

posted by sandy

Suppose we take a random sample of size n from a normal population with variance, σ2 . It can be shown that (n−1)s2/σ2 has a chi-square distribution with n−1 degrees of freedom, where s is the sample variance. Below is a random sample of size 8 drawn from a normal population. Use the above fact to compute a 95% confidence interval for the population variance, σ2 .
Sample 1: 11.6, 17.2, 15.0, 16.3, 22.9, 13.5, 16.4, 16.1

Below is a second random sample, independent from the first, of size 8 from a second normal population. Remembering that the F distribution is a ratio of independent chi- squares divided by their degrees of freedom, it can be shown that, under random, independent sampling, if the variances of the populations are equal, then s21/s2 has an F distribution with, in this case, 7 numerator and 7 denominator degrees of freedom (where the degrees of freedom are n − 1 for the corresponding samples). Test at α = .05 the null hypothesis that the variances are equal against the alternative that the variance of the first population is greater.
Sample 2: 17.7, 11.0, 17.0, 12.4, 10.8, 9.9, 17.2, 10.1

1. MathGuru

Standard Deviation = 3.29 (Standard deviation is the square root of the variance)
Variance = 10.82 (Variance is standard deviation squared)

Using a chi-square table for the endpoints:
(n-1)s^2/16 to (n-1)s^2/1.69
(8-1)10.82/16 to (8-1)10.82/1.69
7(10.82)/16 to 7(10.82)/1.69
75.74/16 to 75.74/1.69
4.73 to 44.82 -->confidence interval for the variance

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For the second problem:
Standard Deviation = 3.43
Variance = 11.76

Sample 1: n = 8; variance = 10.82; df = n - 1 = 7
Sample 2: n = 8; variance = 11.76; df = n - 1 = 7

Test statistic = sample 1 variance / sample 2 variance

You can use the F-distribution at .05 level using the above information for degrees of freedom. This will be your critical value to compare to the test statistic. If the test statistic exceeds the critical value from the table, the null will be rejected in favor of the alternative hypothesis. If the test statistic does not exceed the critical value from the table, then the null is not rejected.

I'll let you take it from here to finish. Check these calculations!

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