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15 of 20 Done Moving to another question will save this rosponse. Question 3 of 8 Question 3 5 points Saved A utility company to compare the of the wind speeds at two proposed stes for their new wind based power plant show the wind speed at she has a normal distribution a random sample of n1 13 wind speed recordings taken at random intervals from site one has a mean speed of xbaf 1 14 2 mph and a standard deviation ofs1 25 mph, and a random sample ofn2 7wind speed recordings taken at random intervals from ste two has a mean speed of ober2 15.8 mph and a standard deviation of s2 52 mph Ir the company wants to test for the equality of the two population variances at 10% significance level, what will be the rejection region(s) at 10% significance level? (Note Ts Test Statistic) TS 3 or T 4 a b TS D 250 or TS 3 c Ts

Explanation / Answer

Using minitab :

Test and CI for Two Variances

Method

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

F method was used. This method is accurate for normal data only.

Statistics

95% CI for
Sample N StDev Variance StDevs
First 13 2.500 6.250 (1.793, 4.127)
Second 7 5.200 27.040 (3.351, 11.451)

Ratio of standard deviations = 0.481
Ratio of variances = 0.231

95% Confidence Intervals

CI for
CI for StDev Variance
Method Ratio Ratio
F (0.208, 0.928) (0.043, 0.862)

Tests

Test
Method DF1 DF2 Statistic P-Value
F 12 6 0.23 0.030

Rejection region F(12,6)=4 ................Critical value using excel function ''=FINV(0.05,12,6)''

So test statistic TS<0.333 and critical region TS>4

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