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In a sample of 8 observations, the sum of the squared deviations of items from their mean was 94.5. in another sample of 10 observations, the value was found to be 101.7. test whether the difference is significant at 5% level.

User Tamilsweet
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The F-test suggests that there is no significant difference in the variances of the two samples at a 5% significance level.

To test the significance of the difference in the variances of two samples, we can use the F-test. The F-statistic is the ratio of the sample variances. The null hypothesis is that the variances are equal, and the alternative hypothesis is that the variances are different.

Let's denote the sum of squared deviations as SS and the degrees of freedom as df. For sample 1 with 8 observations, SS1 is 94.5 and df1 is 7 (8 - 1), and for sample 2 with 10 observations, SS2 is 101.7, and df2 is 9 (10 - 1).

The F-statistic is given by:

F = SS1/df1 divided by SS2/df2

Substitute the values:

F = (94.5/7) divided by (101.7/9)

F is approximately 1.20

Now, compare this F-statistic with the critical values at a 5% level. The critical value for v1 = 3, v2 = 9 is 3.29, and for v1 = 8, v2 = 10 is 3.07.

Since the calculated F-statistic (1.20) is less than both critical values, we fail to reject the null hypothesis. Therefore, there is not enough evidence to suggest that the difference in variances is significant at the 5% level.

The question probable may be:

In a sample of 8 observation, the sum of squared

deviation of item from the mean was 94.5. In

another sample of 10 observation the value was

found to be 101.7. Test whether the difference is

significant at 5% level. You are given that at 5%

level, critical value of F for v1 = 3, V2 = 9 d.f is 3.29

and for V1 = 8, V2 = 10 c.f is 3.07.​

User Zrvan
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