In: Statistics and Probability
Sample 1: n = 38, sample variance = 32.4
Sample 2: n = 44, sample variance = 17.6
We wish to use the F test to test whether the population variances are equal.
a) We will set the null hypothesis
H0 : VS H1 :
b) The F.test statistics is
we are given that S12=32.4, n1=38, S22=17.6 and n2=44,
c) The degree of freedom for numerator and denominator is
DF(Numerator) | DF2(deniminator) |
37 | 43 |
d) We are given the critical value for f test at 0.05 level of significance = 1.685
The calculated f test is given below
Method | DF(Numerator) | DF2(deniminator) | F.Statistic | P-Value |
F Test (normal) | 37 | 43 | 1.84 | 0.055 |
Since the calculated F ratio =1.84 is greater then tabulated F.Ratio =1.685 we accept our null hypothesis and conclude that the two variance are equal. or in other words since calculated P.value=0.055 is greater then 0.05 we accept our null and reject our alternate hypothesis at 0.05 level of significance.
e) MiniTab Output
Test and CI for Two Variances
Method
Null hypothesis Sigma(1) / Sigma(2) = 1
Alternative hypothesis Sigma(1) / Sigma(2) not = 1
Significance level Alpha = 0.05
Statistics
Sample N StDev Variance
1 38 5.692 32.400
2 44 4.195 17.600
Ratio of standard deviations = 1.357
Ratio of variances = 1.841
95% Confidence Intervals
CI for
Distribution CI for StDev Variance
of Data Ratio Ratio
Normal (0.994, 1.867) (0.987, 3.485)
Tests
Test
Method DF1 DF2 Statistic P-Value
F Test (normal) 37 43 1.84 0.055