Question

(Do this problem without using R) Consider the simple linear regression model y =β0 + β1x + ε, where the errors are independent and normally distributed, with mean zero and constant variance σ2. Suppose we observe 4 observations x = (1, 1, −1, −1) and y = (5, 3, 4, 0).

1. (a) Fit the simple linear regression model to this data and report the fitted regression line.

2. (b) Carry out a test of hypotheses using α = 0.05 to determine whether there is a linear relationship between x and y.

3. (c) Carryoutatestofhypothesesusingα=0.05totestH0 :β1 =2VsH1 :β1 ̸=2(This is to make clear that we can do any type of test, not only the special case of β1 = 0).

4. (d) Construct the ANOVA table.

5. (e) Construct 95% CIs for the least-squares estimators and σ2.

6. (f) Construct a 95% CI for the mean response at x = 1.

7. (g) Construct a 95% prediction interval for the future observation at x = 2.

Answer 1

(a) Let the simple linear regression model to this data is y -bo +biT, where bo is y-intercept and b1 is slope Consider the following table Pair_Number xy 25 16 4 SUM 12 4 50 n -50 - (12)-/414 The slope is: The y-intercept is vb 210)3 Therefore, the fitted regression line is

Scatterplot y=x+3 6 4 0 0.5 0 0.5 1.5

(b) The hypotheses are The degrees of freedom is df-n -2 4-2 2 From ttable, at a -0.05 and df- 2, reject null hypothesis if 4.303 Consider the following table y-y (y - g)2 4 4 10 SUM Error sum of squares (Unexplained Variation) tandard error of estimate is SSE 10 4-2 - V5-2.236 The test statistic is 0.894 se/VSS 2.236/V4 Since t0.896< 4.303, do not reject null hypothesis We cannot conclude that there is a linear relationship between x andy

(c) The hypotheses are Ha:B12 The test statistic is bı - 2 =-0.894 se/VSS 2.236/V4 Since 0.896 < 4.303, do not reject null hypothesis. We cannot conclude that slope = 2 Total sum of squares (Total Variation) SST = SS,-14 SSR = SST-SSE = 14 _ 10 = 4. dfR=dfT _dfE = 3-2=1 The ANOVA table is showin below NOVA SS MS-SS/df F-MSreg/MS 4 10 df 0.8 Regression Residual Total 4

(e) Here b1-1, SS,-4 s.-2.236 and critical value of t for df-n-2-4-2-2 at 95% confidence interval for the regression coefficient ist*4.303 2.236 1 4.303 -1 4.81 (-3.81, 5.81) SST (f) AtT1, the predicted value of y is 341 4 The mean of x is 7l The 95% confidence interval for the mean response at x = 1 is = 4 ± 6.80 = (-2.80. 10.80) (g) At 2, the predicted value of y is The 95% confidence interval for future observation at 1-2 is 1(2 0)2 5 4.303(2.236)1/1_ + 7l 5 14.43-9.43, 19.43)