Question

A psychologist is interested in testing whether there is a difference in the distribution of personality types for business majors and social science majors. The results of the study are shown in Table. Conduct a test of homogeneity. Test at a $5\%$level of significance.

$\begin{array}{|llllll|}\hline & \text{Open}& \text{Conscientious}& \text{Extrovert}& \text{Agreeable}& \text{Neurotic}\\ \text{Business}& 41& 52& 46& 61& 58\\ \text{Social Science}& 72& 75& 63& 80& 65\\ \hline\end{array}$

Short answer

Decision: Do not reject the null hypothesis $H_0$

Conclusion: There is sufficient evidence to ensure that the distribution for personality types is the same for business majors and social science majors.

Step-by-step solution

Given Information

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

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 ${\chi }_4^2$.

Tables

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:

$Teststatistic=\sum _{(i\times j)}\frac{(O-E{)}^2}{E}$

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$0.55$

Graph

The Chi-Square sketch is given below: