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.
Answer:
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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