Question

This question is to give you a feel for the actual calculations involved with OLS regressions. In this case, the independent variable is time (t). We will talk more about adding time as a variable later. For now, just treat it like any other independent variable x. You should do the calculations manually without a computer. It is very important that you show your work as answers will vary a bit due to rounding errors so I need to know that you followed the correct methodology. I recommend not trying to type up your answers as typing this many numbers will likely involve at least some typos. Consider the following data (continued on next page):

ti	yi
1	104
2	250
3	310
4	410
5	510
6	610
7	680
8	818
9	943

(e) Find the F-statistic. Is the equation significant at the 1% level? Make sure you state the null and alternative hypotheses. Use the p-value approach.

(f) Find the 95% prediction interval for when t = 5.

(g) There are two ways of calculating SSR (a direct and an indirect way). You would have used one method for previous parts of this question. Now, use the method you did not previously use.

Show your progress and I'll rate your answer good! (It's in Excel btw) Thanks:)

Answer 1

ANSWER:

X	Y	XY	X²	Y²
1	104	104	1	10816
2	250	500	4	62500
3	310	930	9	96100
4	410	1640	16	168100
5	510	2550	25	260100
6	610	3660	36	372100
7	680	4760	49	462400
8	818	6544	64	669124
9	943	8487	81	889249
Ʃx =	Ʃy =	Ʃxy =	Ʃx² =	Ʃy² =
45	4635	29175	285	2990489

Sample size, n =	9
x̅ = Ʃx/n = 45/9 =	5
y̅ = Ʃy/n = 4635/9 =	515
SSxx = Ʃx² - (Ʃx)²/n = 285 - (45)²/9 =	60
SSyy = Ʃy² - (Ʃy)²/n = 2990489 - (4635)²/9 =	603464
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 29175 - (45)(4635)/9 =	6000

e)

Null and alternative hypothesis:

Ho: β₁ = 0

Ha: β₁ ≠ 0

SSE = SSyy -SSxy²/SSxx = 603464 - (6000)²/60 = 3464

SSR = SSxy²/SSxx = (6000)²/60 = 600000

Test statistic:

F = SSR/(SSE/(n-2)) = 600000/(3464/7) = 1212.4711

p-value = F.DIST.RT(1212.4711, 1, 7) = 0.0000

Conclusion:

p-value < α, Reject the null hypothesis.

f)

Predicted value of y at x = 5

ŷ = 15 + (100) * 5 = 515

Significance level, α = 0.05

Critical value, t_c = T.INV.2T(0.05, 7) = 2.3646

95% Prediction interval :

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

= 515 - 2.3646*22.2454*√(1 + (1/9) + ((5 - 5)²/(60))) = 459.5526

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

= 515 + 2.3646*22.2454*√(1 + (1/9) + ((5 - 5)²/(60))) = 570.4474