Question

A straight line is fitted to some data using least squares. Summary statistics are below.

n=10, I = 5, y =12, SSxx=139, SSxy=128, SSyy=155 The least squares intercept and slope are 7.40 and 0.92, respectively, and the ANOVA table is below. Source DF SSMS Regression 1 117.87 117.87 Residual 8 37.13 4.64 Total ||155 [2 pt(s)] What is the estimated mean value for y when x=8? Submit Answer Tries 0/3 [2 pt(s)] If the fitted value for y is 20, what is the corresponding value for x? Submit Answer Tries 0/3 [2 pt(s)] What proportion of the variation in y is explained by the regression in x? Submit Answer Tries 0/3 [2 pt(s)] What is the standard error of the slope estimate? Submit Answer Tries 0/3 To test whether there is any association between X and Y, what are the null and alternative hypotheses? OHO: B1 * 0, HA: B=0 OHO: B2=0, HA: B1 0 HO: B70, HA: B1 = 0 OHO: B1=0, HA: B1 0 [1 pt(s)] Submit Answer Tries 0/1 (3 pt(s)] Test whether there is a significant linear association between X Y. First, compute the value of the test statistic (based on B1). Submit Answer Tries 0/3

Answer 1

The estimated regression line is
y = 7.4 + 0.92x

For x = 8,
y = 7.4 + 0.92 * 8 = 14.76

For y = 20
20 = 7.4 + 0.92x
x = (20 - 7.4) / 0.92 = 13.69565

Proportion of explained variation = SS Regression / SS Total = 117.87 / 155 = 0.7604516

Standard error, Se = $\sqrt{MS~Residual} = \sqrt{4.64}$ = 2.154066

Null and alternative hypotheses are

$H_0: \beta_1 = 0 ; H_a: \beta_1 \ne 0$ (Last Option)

Standard error of slope se(b1) = Se / $\sqrt{SS_{xx}}$ = 2.154066 / $\sqrt{139}$ = 0.1827055

test statistic, t = Coeff / se(b1) = 0.92 / 0.1827055 = 5.035426

Degree of freedom = n-2 = 10 - 2 = 8

P-value = 2 * P(t > 5.035426) = 0.001007339

0.001 < p-value $\le$ 0.02

Critical value of t at df = 8 and 95% confidence interval is  2.306

For x = 9,
yp = 7.4 + 0.92 * 9 = 15.68

95% confidence interval is

yp $\pm$ Se * t * $\sqrt{(1/n) + (x_p - \bar{x})^2/SS_{xx}}$

15.68 $\pm$ 2.154066 * 2.306 * $\sqrt{(1/10) + (9 - 5)^2/139}$

15.68 $\pm$   2.303809

Lower Bound = 15.68 - 2.303809 = 13.37619

Upper Bound = 15.68 + 2.303809 = 17.98381

95% prediction interval is

yp $\pm$ Se * t * $\sqrt{1 + (1/n) + (x_p - \bar{x})^2/SS_{xx}}$

15.68 $\pm$ 2.154066 * 2.306 * $\sqrt{1 + (1/10) + (9 - 5)^2/139}$

15.68 $\pm$   5.475523

Lower Bound = 15.68 - 5.475523 = 10.20448

Upper Bound = 15.68 + 5.475523 = 21.15552