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The null hypothesis is shown below:

$H_0$: The distribution for personality types is same for business majors and social science majors.

Against the alternative hypothesis as shown below:

$H_a$: The distribution for personality types is not same for business majors and social science majors.

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

$df=(numberofcolumns-1)$

Therefore,

$df=(numberofcolumns-1)$

$=5-1$

$=4$

From above calculation, it is clear that the distribution for the test is ${\mathrm{χ}}_4^2$.

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 $\left(E\right)$values table is shown below:

The test statistic of independence test is given below:

$Teststatistic=\underset{(i×j)}{∑}\frac{(O-E{)}^2}{E}$

To calculate $\frac{(O-E{)}^2}{E}$apply formula $=\left((B4-B11{)}^2\right)/B11$in cell B18 and drag the same formula up to cell F19. After that, take total of columns total and rows total.

The table of test statistic is shown below:

Hence, the test statistic is $3.005$

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

Hence, the$p$-value is$0.55$

The Chi-Square sketch is given below: