Question

Below is a regression using X = average price, Y = units sold, n = 20 stores.

R2	0.200
Std. Error	26.128
n	20

ANOVA table
Source	SS	df	MS	F	p-value
Regression	3,080.89	1	3,080.89	4.51	.0478
Residual	12,288.31	18	682.68
Total	15,369.20	19

Regression output					confidence interval
variables	coefficients	std. error	t (df = 18)	p-value	95% lower	95% upper
Intercept	614.9300	51.2343	12.002	.0000	507.2908	722.5692
Slope	−109.1120	51.3623	−2.124	.0478	−217.0202	−1.2038

(a)	Write the fitted regression equation. (Round your answer to 3 decimal places. Negative values should NOT be indicated by a minus sign.)

  YˆY^ =  −  X

(b)	Write the formula for each t statistic and verify the t statistics shown below. (Round your answer to 3 decimal places. Negative values should be indicated by a minus sign.)

	t
Intercept
Slope

(c)	State the degrees of freedom for the t tests and find the two-tail critical value for t by using Appendix D. (Round t.025 value to 3 decimal places.)

df
t.025	±

(d)	Use Excel's function =T.DIST.2T(t, d.f.) to verify the p-value shown for each t statistic (slope, intercept). (Round your answer to 4 decimal places.)

	p-value
Intercept
Slope

Answer 1

a)	Write the fitted regression equation. (Round your answer to 3 decimal places. Negative values should NOT be indicated by a minus sign.)

y^=614.930-109.112x

SolutionB:

	t
Intercept	12.002
slope	-2.124

Solutionc:

df=n-2=20-218

t critical

=T.INV.2T(0.025;18)=

±2.445

df	18
t .025	2.445

Solutiond:

intercept	5.03299E-10=0.0000
slope	0.047785003=0.0478

we gto p value for intercept as

=T.DIST.2T(12.002;18)=

5.03299E-10

we got p value ofr slope as

=T.DIST.2T(2.124;18)

=

0.047785003