Question

An automobile dealer wants to see if there is a relationship between monthly sales and the interest rate. A random sample of four months was taken. Here are the results of the sample

y	x
Monthly Sales	Interest Rate %
22	9.2
20	7.6
10	10.4
45	5.3

The estimated least squares regression equation is: Ŷ= 75.061 ‐ 6.254X

a. Obtain a measure of how well the estimated regression line fits the data.

b. You want to test to see if there is a significant relationship between the interest rate and monthly sales at the 1% level of significance. State the null and alternative hypotheses, and test the hypotheses.

c. Construct a 99% confidence interval for the average monthly sales for all months with a 10% interest rate

d. Construct a 99% prediction interval for the monthly sales of one month with a 10% interest rate.

Please DO NOT use minitab or any software

Answer 1

X	Y	XY	X²	Y²
9.2	22	202.4	84.64	484
7.6	20	152	57.76	400
10.4	10	104	108.16	100
5.3	45	238.5	28.09	2025

Ʃx =	32.5
Ʃy =	97
Ʃxy =	696.9
Ʃx² =	278.65
Ʃy² =	3009
Sample size, n =	4
x̅ = Ʃx/n = 32.5/4 =	8.125
y̅ = Ʃy/n = 97/4 =	24.25
SSxx = Ʃx² - (Ʃx)²/n = 278.65 - (32.5)²/4 =	14.5875
SSyy = Ʃy² - (Ʃy)²/n = 3009 - (97)²/4 =	656.75
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 696.9 - (32.5)(97)/4 =	-91.225

Slope, b = SSxy/SSxx = -91.225/14.5875 = -6.253642

y-intercept, a = y̅ -b* x̅ = 24.25 - (-6.25364)*8.125 = 75.06084

Regression equation :

ŷ = 75.0608 + (-6.2536) x

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1)

Null and alternative hypothesis:

Ho: β₁ = 0

Ha: β₁ ≠ 0

SSE = SSyy -SSxy²/SSxx = 656.75 - (-91.225)²/14.5875 = 86.26

SSR = SSxy²/SSxx = (-91.225)²/14.5875 = 570.4885

Test statistic:

F = SSR/(SSE/(n-2)) = 570.4885/(86.2615/2) = 13.2270

p-value = F.DIST.RT(13.227, 1, 2) = 0.0680

Conclusion:

p-value > α , Fail to reject the null hypothesis.

The estimated regression line does not fits the data well.

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2)

Standard error, se = √(SSE/(n-2)) = √(86.26153/(4-2)) = 6.5674

Null and alternative hypothesis:

Ho: β₁ = 0

Ha: β₁ ≠ 0

Test statistic:

t = b/(se/√SSxx) = -3.6369

df = n-2 = 2

p-value = T.DIST.2T(ABS(-3.6369), 2) = 0.0680

Conclusion:

p-value > α , Fail to reject the null hypothesis.

There is not a significant relationship between the interest rate and monthly sales at the 1% level of significance.

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3)

Predicted value of y at x = 10

ŷ = 75.0608 + (-6.2536) * 10 = 12.5244

Critical value, t_c = T.INV.2T(0.01, 2) = 9.9248

99% Confidence interval :

Lower limit = ŷ - tc*se*√((1/n) + ((x-x̅)²/(SSxx)))

= 12.5244 - 9.9248*6.5674*√((1/4) + ((10 - 8.125)²/(14.5875))) = -33.1485

Upper limit = ŷ + tc*se*√((1/n) + ((x-x̅)²/(SSxx)))

= 12.5244 + 9.9248*6.5674*√((1/4) + ((10 - 8.125)²/(14.5875))) = 58.1974

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4)

99% Prediction interval :

Lower limit = ŷ - tc*se*√(1 + (1/n) + ((x-x̅)²/(SSxx)))

= 12.5244 - 9.9248*6.5674*√(1 + (1/4) + ((10 - 8.125)²/(14.5875))) = -67.0652

Upper limit = ŷ + tc*se*√(1 + (1/n) + ((x-x̅)²/(SSxx)))

= 12.5244 + 9.9248*6.5674*√(1 + (1/4) + ((10 - 8.125)²/(14.5875))) = 92.1140