Question

The following information applies to Problems 14-21. Sup pose an experiment yielded Y values of 1, 3, and 2 when the independent variable X was set to 1, 2, and 3, respectively. 14. Determine the correlation coefficient for these data (E) None of these (A) -0.5 (B) -0.25 (C) 0.25 (D) 0.5 15. Which of the following is an equation of the least-squares regression line? (A) y = 2 (B) y = x (C) y = 1 + 0.5x (D) y- -2+2r (E) None of these 16. Which of the following is the total sum of squares ? (A) 0.5 (B)1 (C) 1.5 (D) 2 (E) None of these )2? 17. Which of the following is the regression sum of squares X(j, -y (A) 0.5 (B)1 (C) 1.5 (D) 2 (E) None of these 18. Which of the following is the coefficient of determination? 19. Determine the error variance estimate s2 (A) 0.5 (B) 1 (C) 1.5 (D) 2 (E) None of these 20. To check if there is a significant linear relation of Y and X, a reseracher carries out a standard t-test for Ho B0 versus Hi P1 0 based on these data, where ^i is the true slope of the assumed regression line. What is the conclusion of the researcher? (A) Reject Ho (B) Do not reject Ho

Answer 1

The regression equation that we want to estimate is

$Y=\beta_0+\beta_1X+\epsilon$

where $\beta_0$ is the intercept of the regression line

is the slope

61

$\epsilon \stackrel{iid}\sim \mathcal{N}(0,\sigma^2)$ is a random disturbance

We calculate the following

n=3 is the sample size

sample means of X and Y are

Σ.Tİ 1+2+3 2

2

The sum pf squares

ss, Σ(zi-i)2 = (1-2)" + (2-2)" + (3-2)2-2 ss, = Σ(Vi-y)2-(1-2)2 + (3-2)2 + (2-2)2-2 SSr,-〉 '(zi-x)(yi-y) = (1-2)(1-2) + (2-2)(3-2) + (3-2)(2-2) = 1

14. The correlation coefficient is

SSr -0.5

ans: D. 0.5

15. The estimate of slope coefficient is

SST 2 31 0.5

The estimate of intercept is

$\begin{align*} \hat{\beta}_0=\bar{y}-\hat{\beta}_1\bar{x}=2-0.5\times 2=1 \end{align*}$

The estimated least square line is

Y = 1 + 0.5X

ans: C. y=1+0.5x

16. We have already calculated the total sum of squares as

$\begin{align*} SS_y=\sum (y_i-\bar{y})^2=2 \end{align*}$

ans: D) 2

17. The regression sum of squares is

$\begin{align*} SSR=\sum (\hat{y}_i-\bar{y})^2=\hat{\beta}_1SS_{xy}=0.5\times 1=0.5 \end{align*}$

ans: A) 0.5

18. The coefficient of determination is

2 SSR 0.5 SST 0.25

ans: A) 1/4

19. The sum of square Error is

SSE = SST-SSR = 2-0.5 = 1.5

The degrees of freedom of error is n-2=3-2=1

Hence the mean square error is

$\begin{align*}MSE =\frac{SSE}{n-2}=\frac{1.5}{3-2}=1.5 \end{align*}$

The estimate of error variance is

$\begin{align*}s^2=MSE =1.5 \end{align*}$

ans: C) 1.5

20. The hypotheses that needs to be tested are

$\begin{align*} &H_0:\beta_1=0\leftarrow\text{null hypothesis: No linear relationship between X and Y}\\ &H_a:\beta_1\ne 0\leftarrow\text{alternative hypothesis: Significant linear relationship between X and Y}\\ &\alpha=0.05\leftarrow\text{level of significance to test the hypothesis: Assumed} \end{align*}$

We know the following

is the estimate of slope

10.5

$\begin{align*} s_{\hat{\beta}_1}=\sqrt\frac{s^2}{SS_x}=\sqrt\frac{1.5}{2}=0.8660 \end{align*}$ is the standard error of slope coefficient

$\begin{align*} \beta_{1H_0}=0 \end{align*}$ is the hypothesized value of the slope coefficient (from the null hypothesis)

The test statistics is

$\begin{align*} t=\frac{\hat{\beta}_1-\beta_{1H_0}}{s_{\hat{\beta}_1}}=\frac{0.5-0}{0.8660}=0.577 \end{align*}$

The degrees of freedom for t statistics is n-2 =3-2=1

This is a 2 tailed test. From the t table for df=1, and level of significance alpha=0.05 (the area under both the tails) we get the critical value = 12.706. That means the acceptance region is -12.706 to +12.706.

We will reject the null hypothesis if the test statistics lies outside the acceptance region.

Here the test statistics is 0.577 and it lies within   -12.706 to +12.706. Hence we do not reject the null hypothesis.

We conclude that there is no sufficient evidence to support the claim that there is significant linear relationship between X and Y.

ans: B) Do not reject null

21. Without calculating we know that the prediction interval for the future Y value is wider than the confidence interval for average Y value. Here we see that (-6.985,10.985) has a width of (10.985-(-6.985)) = 17.97. The width of (-15.969,19.969) is (19.969-(-15.969)) = 35.938
Hence we can say that the prediction interval is (-15.969,19.969)

ans: B) (-15.969,19.969)

Calculations:

The predicted value of Y for X=2 is

$\begin{align*} \hat{y}=1+0.5\times 2=2 \end{align*}$

95% confidence interval indicates a significance level of
a-1-95/1000.05
. The degrees of freedom for t is n-2=3-2=1. Using the t table for df=1 and area under right tail = 0.025 or area under both the tails=0.05 we get the critical value as $\begin{align*} t_{\alpha/2}=12.706 \end{align*}$

The 95% confidence interval for the average Y value is

$\begin{align*} &\hat{y}\pm t_{\alpha/2}s\sqrt{\frac{1}{n}+\frac{(x-\bar{x})^2}{SS_x}}\\ \implies &2\pm 12.706\times \sqrt{1.5}\sqrt{\frac{1}{3}+\frac{(2-2)^2}{2}}\\ \implies &[-6.984,10.984] \end{align*}$

The 95% prediction interval for the future Y value is

$\begin{align*} &\hat{y}\pm t_{\alpha/2}s\sqrt{1+\frac{1}{n}+\frac{(x-\bar{x})^2}{SS_x}}\\ \implies &2\pm 12.706\times \sqrt{1.5}\sqrt{1+\frac{1}{3}+\frac{(2-2)^2}{2}}\\ \implies &[-15.969,19.969] \end{align*}$