Question

A regression analysis of 32 months’ data relating a company's monthly advertising expenses (x, in thousands of dollars) to its sales (y, in thousands of dollars) yields the following output: • ?0=100 • ?1=5.3 • Standard error of the estimate ? = ?? = 56 • Standard error for ?1, ???1 =0.3 Furthermore, when ? ∗=9, the standard error for a confidence interval for the estimated mean response is given by ???̂ = 29, while the standard error for a prediction interval is ???̂ = 63.1.

(a) (3 pts) Is the regression significant at a 5% level of significance?

(b) (3 pts) Say that the company spends $9000 on advertising in a given month. What would you expect their sales to be?

(c) (3 pts) Find a 95% confidence interval for the average sales over all months in which they plan to spend $9000 on advertising.

(d) (3 pts) Assume that you know that in January of 2020, they spent $9000 on advertising, but you don’t have data on their sales. Find a 95% confidence interval for their predicted sales.

Answer 1

a)

Null and alternative hypothesis:

Ho: β₁ = 0 ; Ha: β₁ ≠ 0

n = 32

α = 0.05

Slope, b1 = 5.3

Standard error of slope, se(b1) = 0.3

Test statistic:

t = b1/se(b1) = 5.3/0.3 = 17.6667

df = n-2 = 30

p-value = T.DIST.2T(ABS(17.6667), 30) = 0.0000

Conclusion:

p-value < α Reject the null hypothesis.

Yes, The regression is significant at a 5% level of significance.

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b)

Predicted value of y at x = 9

ŷ = 100 + (5.3) * 9 = 147.7

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c)

Critical value, t_c = T.INV.2T(0.05, 30) = 2.0423

95% Confidence interval for the average sales over all months in which they plan to spend $9000 on advertising:

Lower limit = ŷ - tc*se_?̂ = 147.7 - 2.0423*29 = 88.4741

Upper limit = ŷ + tc*se_?̂ = 147.7 + 2.0423*29 = 206.9259

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d)

95% Prediction interval :

Lower limit = ŷ - tc*se_?̂ = 147.7 - 2.0423*63.1 = 18.8326

Upper limit = ŷ + tc*se_?̂ = 147.7 + 2.0423*63.1 = 276.5674