Question

The data below are the temperatures on randomly chosen days during a summer class and the number of absences on those days. Construct a 95% prediction interval for y, the number of days absent, given x = 95 degrees and y = 0.449x - 30.27

Temperature X	72	86	91	90	88	98	75	100	80
Number of absences Y	3	7	10	10	8	15	4	15	5

a. (9.957, 14.813)

b. (11.378, 13.392)

c. (4.321, 6.913)

d. (3.176, 5.341)

Answer 1

X	Y	XY	X²	Y²
72	3	216	5184	9
86	7	602	7396	49
91	10	910	8281	100
90	10	900	8100	100
88	8	704	7744	64
98	15	1470	9604	225
75	4	300	5625	16
100	15	1500	10000	225
80	5	400	6400	25
Ʃx =	Ʃy =	Ʃxy =	Ʃx² =	Ʃy² =
780	77	7002	68334	813

Sample size, n =	9
x̅ = Ʃx/n =	86.6667
y̅ = Ʃy/n =	8.55556
SSxx = Ʃx² - (Ʃx)²/n =	734
SSyy = Ʃy² - (Ʃy)²/n =	154.222
SSxy = Ʃxy - (Ʃx)(Ʃy)/n =	328.667

Slope, b = SSxy/SSxx = 0.44777

y-intercept, a = y̅ -b* x̅ = -30.2516

Regression equation :

ŷ = -30.2516 + (0.4478) x

Predicted value of y at x =95

ŷ = -30.2516 + (0.4478) * 95 = 12.287

df = n-2 = 7

Significance level, α = 0.05

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

Sum of Square error, SSE = SSyy -SSxy²/SSxx = 7.05359

Standard error, se = √(SSE/(n-2)) = 1.00382

95% prediction interval for y

Answer a)

Note: The answer is little different because of the rounding off of numbers.