In: Statistics and Probability
Consider the following measures based on independently drawn samples 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 the population variances. (Round "F" value and final answers to 2 decimal places.)
Confidence interval _____ to _____
b. Using the confidence interval from Part a, test if the ratio of the population variances differs from one at the 5% significance level. Explain.
The 95% confidence interval (does not contain / contains) the value 1.
Thus, we (can, cannot) conclude that the population variances differ at the 5% significance level.
I assume that 249 and 236 are sample variance and not sample standard deviation
a)
Test and CI for Two Variances
Method
Null hypothesis
σ(First) / σ(Second) = 1
Alternative hypothesis σ(First) / σ(Second) ≠ 1
Significance level α = 0.05
F method was used. This method is accurate for normal data only.
Statistics
95% CI for
Sample N StDev
Variance StDevs
First 51 15.780 249.000
(13.203, 19.615)
Second 26 15.362 236.000 (12.048, 21.206)
Ratio of standard deviations = 1.027
Ratio of variances = 1.055
95% Confidence Intervals
CI for
CI for
StDev Variance
Method
Ratio
Ratio
F (0.712, 1.423) (0.508,
2.024)
Tests
Test
Method DF1 DF2 Statistic P-Value
F 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 population variances differ at the 5% significance level
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