1. We have interest in the determinants of college GPA. The table below shows th

Question

1. We have interest in the determinants of college GPA. The table below shows the regression's results Source df MS Number of obs Model Residual 295.60292 4128 9.14802161 071609235 Prob> F R-squared Adj R-squaredX.xxxx Root MSE 54.88813 6 XXxXX 0.1566 Total 350.49105 4134 084782547 0.2676 lcol oe Std. Err. t PPt 95% Conf. Interval lsat hsize hsizesq female 0.7633204 0.0094815 0.00273 0.0957847 0.052844 031610 xxxx 0077148XxxX 0010606 xxx 008716 xxxx 0261649 xxxx 0364934 xxxx 0.7013475 0.825293 xxxxx -0.0056438 0.024607 xxxxx 0.0048106 -0.000652 0.0786965 0.112873 0.1041415 -0.001547 xxxxx 0.1177314 0.025362 4.814835 -3.954995 ac femaleblack -0.046185 cons 4.384915 2192868 xxxx where lcolgpa, is the log of students GPA after fall semester, Isat is the log of SAT scores, hsize is is the size of the student's high school graduating class, in hundreds, female is a gender dummy variable, and black is a race dummy variable a. Clearly state the null and alternative hypotheses to test whether a 1% increase in SAT result in a change different than 1% in GPA b, Using the critical value approach, conduct the hypothesis test at 1% level of significance hat is your conclusion? c. Compute the one-sided p-value for hsize d. Holding everything constant, what would be the average change in GPA when we compare the size of high school graduating class A (size 300) and B (size 500)?

Explanation / Answer

Against the,

ALTERNATIVE HYPOTHESIS: A one per cent INCREASE in the SAT result causes a change different than 1% (more or less than 1%) in the GPA. “Change in GPA” 0

Standard error for the SAT score =0.0316101

Estimate of the coefficient is = 0.7633204

Now, we know that the t-statistic is given by dividing the estimate of the parameter by the standard error i.e

     t = Parameter Estimate/Standard error of the parameter.

Putting the given values we have

t = 0.7633204/0.0316101

t = 24.14799067

The degrees of freedom for this statistic is n-k-1 i.e 4134

The critical value for a two tailed test is±2.80854182

Here we notice that the value of the t-statistic is much greater than the critical value of the t-distribution for the given degrees of freedom.

t-statistic > t-critical

So, we reject our null hypothese and hence we conclude that a 1% increase in the SAT score causes more than 1% change in the GPA.

T-statistic = 24.14799067

Degrees of freedom = 4134

Using an online p-value calculator, we get the p-value as = 0.0001.

This result is extremely statistically significant.

We reject the null hypothesis at 1% level of significance.

Y = a + bX

Where Y is the GPA score

X is the value of the class size

b is the estimate of the coefficient given as 0.0094815

a is the constant term accounting for all the other factors affecting the GPA which are held constant.

So, our regression equation is Y= a + 0.0094815X

So a the average change in the GPA can be obtained as follows

At class size of 300, Y1 = a + 0.0094815x300 = a +2.84445

At class size of 400, Y2 = a +0.0094815x400 = a + 3.7926

So, the change is given by Y2 –Y1 =(a+3.7926)-(a+2.84445) =0.94815

Thus, on an average the GPA changes by 0.94815 if the class size increases by 100.

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