Aim: To test the sample values are consistent with the population variance 64.
H0: The population variance s2 = 64 for the exam scores.
H1: The population variance s2 ¹ 64 for the exam scores.
Basic Situation and assumptions:
To test the variance of single sample of 25 exam scores. The sample are assumed as random and from Normal population N (m,s2) with population Mean m and population variance s2. S2 is the sample variance and is the Sample mean.
The appropriate test statistic to test a single sample variance is Chi-square test.
~ under H0.
Here n = sample size, S2 = Sample variance and is the variance to be tested.
Level of Significance: The level of significance of the test is 5%
Calculation of Sample Mean and Sample Variance:
Sample Variance = S2 = and sample mean = =
= (80+79+69+71+74+73+77+75+65+52+81+84+84+79+70+78+62+77+88+70+75+85+84) / 25
= 1877 / 25
Hence the Sample Mean of exam scores = 75.08
We can work out the alternate formula for easy calculation as follows
3 xi2 = 802+792+692+712+742+732+772+752+652+522+812+842+842+792+702+782+622+772+882+702+ 752+852+842 = 142505
Hence substituting the values in the test statistic
This is a two tailed test since the variance may be less than or greater than 64. At 5% significance level of significance we have to look for the upper 2.5% and lower 2.5% points of the Chi-square distribution table with 24 degrees of freedom.
The lower 2.5% and upper 2.5% points of are 12.4 and 39.36.
Our test statistic value 24.685 lies in the above interval (12.4, 39.36). Hence at 5% level of significance we have insufficient evidence to reject the null hypothesis and also the p-value for the test statistic is 0.8461 which is large enough than 0.05 and hence H0 withstands.
As we have insufficient evidence to reject the null hypothesis, we can conclude that the sample values of 25 exam scores are consistent with the true variance 64.
Critical points of Chi-square distribution.
The Mega stat analysis in Excel 2003.