Question

The following data represent the commute time​ (in minutes) x and a score on a​ well-being survey y. The equation of the​ least-squares regression line is Modifying Above y with caret equals negative 0.0512 x plus 70.2089 and the standard error of the estimate is 0.2843. Complete parts ​(a) through ​(e) below.

   x 25 35 45 55 70 92 125    y 69.0 68.7 67.6 67.2 66.5 65.9 63.7 ​

(a) Predict the mean​ well-being index composite score of all individuals whose commute time is 30 minutes. ModifyingAbove y with caret equals 68.67 ​(Round to two decimal places as​ needed.)

​(b) Construct a​ 90% confidence interval for the mean​ well-being index composite score of all individuals whose commute time is 30 minutes. Lower Bound equals nothing ​ (Round to two decimal places as​ needed.) Upper Bound equals nothing ​ (Round to two decimal places as​ needed.)

c) Predict the well being composite score of Jane whose commute time is 30 mins

d) Construct a​ 90% prediction interval for the well-being index composite score of Jane whose commute time is 30 minutes. Lower Bound equals nothing ​ (Round to two decimal places as​ needed.) Upper Bound equals nothing ​ (Round to two decimal places as​ needed.)

Need help with b-d. Thank you!

Answer 1

X	Y	XY	X²	Y²
25	69	1725	625	4761
35	68.7	2404.5	1225	4719.69
45	67.6	3042	2025	4569.76
55	67.2	3696	3025	4515.84
70	66.5	4655	4900	4422.25
92	65.9	6062.8	8464	4342.81
125	63.7	7962.5	15625	4057.69
Ʃx =	Ʃy =	Ʃxy =	Ʃx² =	Ʃy² =
447	468.6	29547.8	35889	31389

Sample size, n =	7
x̅ = Ʃx/n = 447/7 =	63.8571429
y̅ = Ʃy/n = 468.6/7 =	66.9428571
SSxx = Ʃx² - (Ʃx)²/n = 35889 - (447)²/7 =	7344.85714
SSyy = Ʃy² - (Ʃy)²/n = 31389.04 - (468.6)²/7 =	19.6171429
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 29547.8 - (447)(468.6)/7 =	-375.657143

Slope, b = SSxy/SSxx = -375.65714/7344.85714 =    -0.051145602

y-intercept, a = y̅ -b* x̅ = 66.94286 - (-0.05115)*63.85714 =    70.20886918

Regression equation :   

ŷ = 70.2089 + (-0.0511) x  

a) Predicted value of y at x =    30

ŷ = 70.2089 + (-0.0511) * 30 = 68.67

b) Significance level, α =    0.1

Critical value, t_c = T.INV.2T(0.1, 5) = 2.015  

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

= 19.61714 - (-375.65714)²/7344.85714 =    0.403932003

Standard error, se = √(SSE/(n-2)) = √(0.40393/(7-2)) =    0.28423

90% Confidence interval :

c)

Predicted value of y at x =    30

ŷ = 70.2089 + (-0.0511) * 30 = 68.67

d) 90% prediction interval :