Chapter 15, Problem 29CAP

1.

To determine

Find the value of point-biserial correlation coefficient.

Answer to Problem 29CAP

The value of point-biserial correlation coefficient is 0.396.

Explanation of Solution

Calculation:

The given information is that, the researcher is interested to measure the relation between the factors extracurricular activity and college freshman GPA.

Point-biserial correlation coefficient:

When the one factor is continuous that is measured on either interval scale or ratio scale and other is dichotomous that is measured on nominal scale, then the strength and direction of the linear relationship between the two factors can be measured by Point-biserial correlation coefficient. It is denoted by $r_{\text{pb}}$. The formula is,

$r_{\text{pb}}=\left(\frac{M_{Y_1}-M_{Y_2}}{s_Y}\right)\left(\sqrt{pq}\right)$

In the formula, p and q are the proportion of scores for the dichotomous factor at each level, $s_Y=\sqrt{\frac{S{S}_Y}{n}}$ denotes the standard deviation of Y scores, n denotes the number of pairs measured, $S{S}_Y$ denotes the sums of squares for scores Y, $M_{Y_1}$ denotes the mean of Y scores at first level, $M_{Y_2}$ denotes the mean of Y scores at second level.

The Point-biserial correlation for extracurricular activity and college freshman GPA is to be calculated. The factor ‘extracurricular activity’ has to be coded with 1 and 2. Code the value 1 for ‘No’ and 2 for ‘Yes’ for dichotomous factor ‘extracurricular activity’.

The coded data is,

Extracurricular activity	College Freshman GPA
2	3.52
2	3.33
2	3.91
2	3.70
1	2.96
1	3.84
1	3.42
1	2.76
1	3.88
1	2.80

Software procedure:

Step by step procedure to obtain test statistic value for extracurricular activity and college freshman GPA using SPSS software is given as,

• Choose Variable view.
• Under the name, enter the name as Activity, GPA.
• Choose Data view, enter the data.
• Choose Analyze>Correlate>Bivariate.
• In variables, enter the Activity, and GPA.
• Click OK.

Output using SPSS software is given below:

The correlation coefficient value is 0.396.

2.

To determine

Determine the decision for the null hypothesis using two-tailed test at a 0.05 level of significance

Answer to Problem 29CAP

The decision is to retain the null hypothesis.

The point-biserial correlation coefficient between extracurricular activity and college freshman GPA is significant.

Explanation of Solution

Calculation:

The given information is that, correlation coefficient value is 0.396 and degrees of freedom are 8. The correlation between extracurricular activity and college freshman GPA is to be tested.

Test statistic:

The coefficient of determination for the point-serial correlation and the effect size of the two-independent-sample t test are similar and hence for significance test of point-biserial correlation, t test is used.

The formula for converting the r to a t statistic is,

$t^2=\frac{r^2}{\left(\frac{\left(1-r^2\right)}{df}\right)}$

In the formula, r denotes the correlation coefficient and df is the degrees of freedom for correlation coefficient.

Degrees of freedom:

The formula is. $df=n-2$.

In formula, n denotes the sample size.

Decision rules:

• If the test statistic value is greater than the critical value, then reject the null hypothesis and test is significant, or else retain the null hypothesis.
• If the negative test statistic value is less than negative critical value, then reject the null hypothesis and test is significant, or else retain the null hypothesis.

Let $\rho$ denote the correlation coefficient between extracurricular activity and college freshman GPA.

Null hypothesis:

$H_0:\rho =0$

That is, the point-biserial correlation coefficient between extracurricular activity and college freshman GPA is not significant.

Alternative hypothesis:

$H_1:\rho \ne 0$

That is, the point-biserial correlation coefficient between extracurricular activity and college freshman GPA is significant.

Degrees of freedom:

A sample of 10 participants is taken. The degrees of freedom are,

$\begin{array}{c}df=n-2\\ =10-2\\ =8\end{array}$

Test statistic:

Substitute, $r=0.396$ and degrees of freedom as 8 in test statistic formula

$\begin{array}{c}{t}^2=\frac{{\left(0.396\right)}^2}{\left(\frac{1-{\left(0.396\right)}^2}{8}\right)}\\ =\frac{0.1568}{\left(\frac{1-0.1568}{8}\right)}\\ =\frac{0.1568}{0.1054}\\ =1.4877\end{array}$

The value of t is,

$\begin{array}{c}t=\sqrt{1.4877}\\ =1.219\end{array}$

Critical value:

The given significance level is $\alpha =0.05$.

The test is two tailed, the degrees of freedom are 8, and the alpha level is 0.05.

From the Appendix B: Table B.2 the t Distribution:

• Locate the value 8 in the degrees of freedom (df) column.
• Locate the 0.05 in the proportion in two tails combined row.
• The intersecting value that corresponds to the 8 with level of significance 0.05 is 2.306.

Thus, the critical value for $df=8$ with two-tailed test is $\pm 2.306$.

Conclusion:

The value of test statistic is 1.219.

The critical value is 2.306.

The test statistic value is greater than the critical value.

The test statistic value falls under critical region

Hence the null hypothesis is rejected and test is significant.

Thepoint-biserial correlation coefficient between extracurricular activity and college freshman GPA is not significant.