Question

The summary statistics for the first chemistry test and the second chemistry test scores of 62 students are as follows.

	1st test	2nd test
Sample Mean	88.63	84.44
Sample Standard Deviation	11.46	13.70

The teacher in charge of chemistry assumed a simple regression model that predicts the results of the 2nd test with the results of the 1st test, and obtained the following analysis of variance table through simple regression analysis.

	DF	Sum of Squares	Mean Square	F
Regression
Residuals		1236.1
Total		11460.9

a) Complete the analysis of variance table above.

b) Test the significance of the simple return model at a significant level of 5% using the F test.

c) Test the significance of the slope and explain the relationship to the test for the simple regression model.

d) If your 1st test score is 85, obtain a 95% confidence interval for your average 2nd test score.

Answer 1

a)

through simple regression analysis.

	DF	Sum of Squares	Mean Square	F
Regression	1	10224.8	10224.8	496.3093
Residuals	60	1236.1	20.60167
Total	61	11460.9

k is number of predictors

DF Regression = k = 1

DF Residuals = n-k-1 = 62-1-1 = 60

DF Total = n-1 = 62-1 = 61

SS Regression = SS Total - SS Residuals = 11460.9 - 1236.1 = 10224.8

MS = SS / DF

F = MS Regression / MS Residuals = 10224.8 / 20.60167 = 496.3093

b)

DF for F statistic is k, n-k-1 = 1, 60

Critical value of F at df = 1, 60 and 0.05 significance level is 4.00

Since the observed F (496.3093) is greater than the critical value, we reject the null hypothesis H0 and conclude that simple return model is significant at level of 5%

c)

r = sqrt(SS Regression / SS Total) = sqrt(10224.8 / 11460.9) = 0.944535

Point Estimate of Slope = r Sy / Sx = 0.944535 * (13.70 / 11.46) = 1.129156

Standard error of regression, se = sqrt(MS Residuals) = sqrt(20.60167) = 4.538906

Standard error of slope = se / sqrt((n-1) * S^2x) = 4.538906 / sqrt(61 * 11.46^2) = 0.05071094

Test statistic, t = Slope Coeff / Std Error = 1.129156 / 0.05071094 = 22.26652

Critical value of t at df = 60 and 0.05 significance level is 2.00

Since the observed t (22.26652) is greater than the critical value, we reject the null hypothesis H0 and conclude that slope and the relationship to the test for the simple regression model is significant at level of 5%

d)

Intercept = 84.44 - 1.129156 * 88.63 = -15.6371

= -15.6371 + 1.129156 x

For x = 85,

= -15.6371 + 1.129156 * 85 = 80.34116

95% confidence interval for your average 2nd test score is,

se * t *

80.34116 4.538906 * 2.00 *

80.34116 0.1613476

(80.34116 - 0.1613476 ,  80.34116 + 0.1613476)

(80.17981 , 80.50251)