Question

A research team collected data on n= 450 students in a statistics course. The observed average final examination score was 524, with an observed standard deviation of 127.6 (the divisor in the estimated variance was n-1). The average first examination score was 397, with an observed standard deviation of 96.4. The correlation coefficient between the first examination score and the final examination score was 0.63. a. Report the analysis of variance table and result of the test of the null hypothesis that the slope of the regression line of final exam score on first exam score is zero against the alternative that it is not. Use the 0.10, 0.05, and 0.01 levels of significance. b. Determine the least-squares fitted equation and give the 99% confidence interval for the slope of the regression of final examination score on first examination score.

Answer 1

a.

Let x and y be the first and final examination score.

Given,

$\bar{x}$ = 397, Sx = 96.4

$\bar{y}$ = 524, Sy = 127.6

r = 0.63

SST = (n-1) Sy² = (450 - 1) * 127.6² = 7310510.24

SSR = r² * SST = 0.63² * 7310510.24 = 2901541.51

SSE = SST - SSR = 7310510.24 - 2901541.51 = 4408968.73

DFR = k-1 = 2 - 1 = 1 (k is number of coefficients in the regression model)

DFE = n-k = 450 - 2 = 448

DFT = n-1 = 450 - 1 = 449

MSR = SSR / DFR = 2901541.51 / 1 = 2901541.51

MSE = SSE / DFE = 4408968.73 / 448 = 9841.448

Test statistic, F = MSR / MSE = 2901541.51 / 9841.448 = 294.83

P-value = P(F > 294.83, df = 1, 448) = 0.0000

Anova table is,

Source of variation	SS	DF	MS	F	P-value (Pr>F)
Regression	2901541.51	1	2901541.51	294.83	0.0000
Error	4408968.73	448	9841.448
Total	7310510.24	449

$H_0: \beta_1 = 0$

$H_a: \beta_1 \ne 0$

Estimated coefficient, b1 = r (Sy / Sx) = 0.63 * (127.6 / 96.4) = 0.8339

Standard error of regression, s = $\sqrt{MSE} = \sqrt{9841.448}$ = 99.20407

Sxx = (n-1) Sx² = (450 - 1) *  96.4² = 4172539.04

Standard error of slope coefficient, SE(b1) = s / $\sqrt{S_{xx}}$ = 99.20407 / $\sqrt{4172539.04}$ = 0.04857

Test statistic, t = b1 / SE(b1) = 0.8339 / 0.04857 = 17.169

Degree of freedom = DFE = 448

P-value = 2 * P(t > 17.169, df = 448) = 0.0000

Since, p-value is less than 0.01 significance level, we reject null hypothesis H0 and conclude that there is strong evidence that slope of the regression line of final exam score is not 0 at all given level of significance (0.10, 0.05 and 0.01).

b.

Intercept b0 = $\bar{y}$ - b1 $\bar{x}$ = 524 - 0.8339 * 397 = 192.9417

The least-squares fitted equation is,

y = 192.9417 + 0.8339 x

Critical value of t at 99% confidence interval and df = 448 is  2.587

99% confidence interval of the slope coefficient is b1 $\pm$ t * SE(b1)

(0.8339 - 2.587 * 0.04857, 0.8339 + 2.587 * 0.04857)

(0.7082 ,  0.9596)