# Consider the following measure

Consider the following measures based on independently drawnsamples from normally distributed populations: Use Table 4.

Sample 1: 1formula81.mml = 249, and n1 = 51

Sample 2: 1formula82.mml = 236, and n2 = 26

a. Construct the 95% interval estimate for the ratio of thepopulation variances. (Round “F” value and final answers to 2decimal places.)

Confidence interval _____ to _____

b. Using the confidence interval from Part a, test if the ratioof the population variances differs from one at the 5% significancelevel. Explain.

The 95% confidence interval (does not contain /contains) the value 1.

Thus, we (can, cannot) conclude that the populationvariances differ at the 5% significance level.

I assume that 249 and 236 are sample variance and not samplestandard deviation

a)

Test and CI for Two Variances

Method

Null hypothesis        σ(First) / σ(Second) = 1Alternative hypothesis σ(First) / σ(Second) ≠ 1Significance level      α = 0.05

F method was used. This method is accurate for normal dataonly.

Statistics

95% CI forSample   N   StDevVariance       StDevsFirst      51 15.780   249.000(13.203, 19.615)Second 26 15.362   236.000 (12.048, 21.206)

Ratio of standard deviations = 1.027Ratio of variances = 1.055

95% Confidence Intervals

CI for         CI forStDev      VarianceMethod      Ratio          RatioF       (0.712, 1.423) (0.508,2.024)

Tests

TestMethod DF1 DF2 Statistic P-ValueF          50  25      1.06       0.909

95% interval estimate for the ratio of the population variances.(0.508, 2.024)

b)

The 95% confidence interval (contains) the value 1.

Thus, we (cannot) conclude that the populationvariances differ at the 5% significance level

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