91Ó°ÊÓ

We have to test the energy use values in these two areas come from the same distribution

The null hypothesis is shown below:$H_0:$ : The distribution of energy uses values are the same.

Against the alternative hypothesis as given below:

$H_a:$The distribution of energy uses values are different.

Degrees of freedom can be calculated by the formula given below

$df=(numberofcolumns-1$)

=2-1

=1

From above calculation, it is clear that the distribution for the test is ${\mathrm{χ}}_1^2$.

The observed value table is already given in the textbook. Calculate the expected frequencies by using the formula shown below:

$E=\frac{\left(\text{row total}\right)\left(\text{column total}\right)}{\text{overall total}}$

All calculations can be done in excel worksheet. Hence, the expected $(E)$ values table is shown below:

The test statistic of independence test is given below:

Test statistic$=\underset{(i×f)}{∑}\frac{(O-E{)}^2}{E}$

To calculate $\frac{(O-E{)}^2}{E}$apply formula $=\left((B4-B16{)}^2\right)/B16\mathrm{j}$in cell B27 and drag the same formula up to cell C32. After that, take total of columns total and rows total. The table of test statistic is shown below:

The test statistic is 2.74

The $p-$value can be calculated in excel by using CHIDIST ( ) formula as shown below:

the p- value is 0.098

The Chi-square sketch is given below

T

i. Alpha:$0.05$

ii: Decision: Do not reject the null hypothesis$H_0$

iii. Reason for decision: Because $p$-value $>\mathrm{Î\pm }$

iv. Conclusion: There is sufficient evidence to insure that the distribution of energy uses values are same.