To test the sample values are consistent with the population variance 64 Essay

Aim: To test the sample values are consistent with the population variance 64.

Null Hypothesis:

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H0: The population variance s2 = 64 for the exam scores.

Alternate Hypothesis:

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.

Test Statistic:

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 =  =

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

= 75.08

Hence the Sample Mean of exam scores = 75.08

We can work out the alternate formula for easy calculation as follows

S2 =

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

S2 =

=

Hence substituting the values in the test statistic

=

 

Decision:

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.

Conclusion:

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.

 

 

 

 

 

 

 

 

 

 

 

 

Appendix:

Critical points of Chi-square distribution.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The Mega stat analysis in Excel 2003.