Question

The exercise involving data in this and subsequent sections were designed to be solved using Excel. The following estimated regression equation is based on 10 observations was presented. y = 29.1270 +0.59061 +0.49802, Here SST = 6,532.000, SSR = 6,078.625, 86, = 0.0775, and 8h = 0.0670. a. Compute MSR and MSE (to 3 decimals). MSR MSE b. Compute the F test statistic (to 2 decimals). Use F table. What is the p-value? - Select your answer - At a = .05, what is your conclusion? - Select your answer - c. Compute the t test statistic for the significance of B1 (to 2 decimals). Use t table. The p-value is - Select your answer At a = .05, what is your conclusion?

need it ASAP thanks

Answer 1

(a)

There are 2 predictor variables so degree of freedom of regression is

$df_{reg}=2$

There are total 10 observations so degree of freedom of error is

$df_{res}=10-2-1=7$

$MSR=\frac{SSR}{df_{reg}}=3039.313$

$MSE=\frac{SST-SSR}{df_{res}}=64.768$

(B)

The F test statistics is

$F=\frac{MSR}{MSE}=46.93$

The p-value is

p-value < 0.001

Since p-value is less than level of significance so we reject the null hypothesis.

(C)

Here we have b = 0.5906, n = 10, sb, = 0.0775 Hypotheses are: Ho:B2 = 0 H :B 0 Level of significance: a =0.05 Test is two tailed. Degree of freedom: df =n-2=8 Test Statistics: P1 = (0.5906 - 0)/ 0.0775 sha = 7.62 Critical value: Rejection Region: 2.306 Ift<-2.306 ort > 2.306, Reject HO Decision: Since test statistics lies in rejection region so we reject the null hypothesis P-value: P-value=0 Decision: Since p-value is less than level of significance so we reject the null hypothesis Excel function for critical value: ETINV(0.05,8) Excel function for p-value: ETDIST(7.62,8,2)

Answers:

t = 7.62, p-value < 0.001, Reject H0

(d)

Here we have b = 0.498, n = 10, sb, = 0.067 Hypotheses are: H:B2 = 0 H.:B270 Level of significance: a =0.05 Test is two tailed. Degree of freedom: df =n-2=8 Test Statistics: t-bz-P2 = (0.498 -0)/0.067 - sb₂ = 7.43 Critical value: Rejection Region: 2.306 Ift<-2.306 ort> 2.306, Reject HO Decision: Since test statistics lies in rejection region so we reject the null hypothesis P-value: P-value=0 Decision: Since p-value is less than level of significance so we reject the null hypothesis Excel function for critical value: Excel function for p-value: ETINV(0.05,8) ETDIST(7.43,8,2)

Answers:

t = 7.43, p-value < 0.001, Reject H0