Question

(27 pts total) What follows is hypothetical Excel output for doing regression of y on x. The regression equation is: ^y = 54:6429 + 0:9451x Source DF SS MS F P Regression 1 130.0396 130.0396 20.7812 0.0019 Residual Error Total 9 180.1000 (a) (3 pts) Complete the ANOVA table above by lling in the \Residual Error" row above. (b) (6 pts) What are the values of the regression standard error s and the coecient of determination r2? (c) (6 pts) Write down a null hypothesis H0 and an alternative hypothesis Ha that the P-value 0:0019 in this chart is referring to. Should you reject the null hypothesis at the 1% level of signcance? To what type of error are you now subject to? (d) (4 pts) Suppose SSxx = 145:6. Calculate the t statistic for the test in part (c) and conrm the P-value using the t-table. There are a couple dierent ways this could be done. Choose one, but show your work and/or explain. (e) (8 pts) Suppose x = 20:8 and SSxx = 145:6. Give a 90% prediction interval for the value of y when x = 24. Describe the meaning of this interval in relation to the population

Answer 1

(a)

The complete ANOVA table is,

Source DF SS MS F P

Regression 1 130.0396 130.0396 20.7812 0.0019

Residual Error 8 50.0604 6.25755

Total 9 180.1000

DF for Residual Error = DF Total - DF for Regression = 9 - 1 = 8

SS for Residual Error = SS Total - SS for Regression = 180.1000 - 130.0396 = 50.0604

MS for Residual Error = SS for Residual Error / DF for Residual Error = 50.0604 / 8 = 6.25755

(b)

Regression standard error s =

= 2.50151

Coefficient of determination r2 = SS for Regression / SS for Total = 130.0396 / 180.1000 = 0.722

(c)

where is the coefficient for x in the model.

Since p-value is less than 0.01 significance level, we reject null hypothesis H0 and conclude that there is significant evidence that .

Since we are rejecting the null hypothesis, there may be chance that the null hypothesis is true and we may commit Type I error.

(d)

Standard error of = s / = 2.50151 / = 0.2073106

t statistic = Coeff Estimate / Standard error = 0.9451 / 0.2073106 = 4.56

P-value = 2 * P(t > 4.56, df = 8) = 0.0018

which is approximately equal to P-value 0:0019 in Anova table.

(e)

Given,

= 20.8

SSxx = 145.6

y when x = 24 is,

^y = 54.6429 + 0.9451 * 24 = 77.3253

Critical value of t at 90% confidence level and df = 8 is, 1.86

90% prediction interval for the value of y when x = 24 is,

= (77.3253 - 5.0335,  77.3253 + 5.0335)

= (72.2918, 82.3588)

We are 90% confident that the next new observation (y) will fall within (72.2918, 82.3588) when x is 24.