Question

In a study to determine if response time, y, could be modeled as a linear function of the temperature, a process was run at each of four temperatures three times, a total of twelve pairs of observations. The two AOV tables shown below are (I) for fitting y as a simple linear function of x, and (II) for AOV using the values of x to define the "treatments." Analysis of Regression Sum of Mean Source DF Squares Square Regression 290.40000 290.40000 Error 43.60000 4.36000 Total 334.00000 Analysis of Variance DE Source Groups Error Total Sum of Mean Squares Square 318.0000000 106.0000000 16.0000000 2.0000000 334.0000000

Answer 1

Degrees of freedom can be calculated using mean square and sum of squares.

Mean square = SS/df

df = SS/MS

For Analysis of regression

df regression = SS/MS = 290.4/290.4 = 1

df error = 43.60/4.360 = 10

df total = df regression + df error = 1 + 10 = 11

Source	degrees of freedom	Sum of Squares	Mean square
Regression	1	290.40	290.40
Error	10	43.60	4.36
Total	11	334

Similarly for Analysis of variance

Source	degrees of freedom	Sum of Squares	Mean square
Groups	3	318.0	106.0
Error	8	16.0	2.0
Total	11	334

b)

Hypothesis

H₀ : Slope coefficient = 0

H₁: Slope coefficient $\neq$ 0

The F-statistic for the  Analysis of regression = MS regression / MS Error = 290.4/4.36 = 66.605

considering alpha = 0.05 critical value = F_0.05,1,10 = 4.965

The p-value is < .00001

Since the p-value is less than alpha (0.05) we reject the null hypothesis and conclude that there is significant evidence that the slope coefficient is not equal to zero.

c) From the regression analysis, we get F-statistic = 66.605

critical value for F = 4.965

Since the test statistic is greater than critical value we reject the null hypothesis and conclude that there is significant evidence that straight line does provide an appropriate fit.