91Ó°ÊÓ

Given that the survey results shown in the table is the distribution of applicable reasons the same as the distribution of the most important reason

The null hypothesis is given below:

$H_0$ : The distribution of applicable reasons is the same as the distribution of the most important reason.

Against the alternative hypothesis as given below:

$H_a:$The distribution of applicable reasons is different from the distribution of the most important reason.

The degrees of freedom can be calculated by the formula given below:

$df=(\text{number of columns}-1)$

Therefore.

$df=(\text{number of columns}-1)$

$=2-1$

$=1$

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

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

The test statistic of the independence test is given below

$\text{Test staristic}=\underset{(<1)}{∑}\frac{(O-E{)}^2}{E}$

To calculate $\frac{(O-E{)}^2}{E}$apply the formula

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

Hence the test statistic is234

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

The p-value is 0.000

i. Alpha:$0.05$

ii: Decision: 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 applicable reasons is different as the distribution of the most important reason.