Question

The grades of a class of 9 students on a midterm report (x) and on the final examination (y) are as follows:

  

x

77 50 71 72 81 94 96 99 67

y

82 66 78 34 47 85 99 99 68

Compute a 95% confidence interval for the mean response mu subscript Y divided by x end subscript when x equals 85.

Answer 1

X	Y	XY	X²	Y²
77	82	6314	5929	6724
50	66	3300	2500	4356
71	78	5538	5041	6084
72	34	2448	5184	1156
81	47	3807	6561	2209
94	85	7990	8836	7225
96	99	9504	9216	9801
99	99	9801	9801	9801
67	68	4556	4489	4624
Ʃx =	Ʃy =	Ʃxy =	Ʃx² =	Ʃy² =
707	658	53258	57557	51980

Sample size, n =	9
x̅ = Ʃx/n = 707/9 =	78.55555556
y̅ = Ʃy/n = 658/9 =	73.11111111
SSxx = Ʃx² - (Ʃx)²/n = 57557 - (707)²/9 =	2018.222222
SSyy = Ʃy² - (Ʃy)²/n = 51980 - (658)²/9 =	3872.888889
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 53258 - (707)(658)/9 =	1568.444444

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

= 3872.88889 - (1568.44444)²/2018.22222 = 2653.985466

Standard error, se = √(SSE/(n-2)) = √(2653.98547/(9-2)) = 19.4715

Slope, b = SSxy/SSxx = 1568.44444/2018.22222 = 0.777141599

y-intercept, a = y̅ -b* x̅ = 73.11111 - (0.77714)*78.55556 =    12.06232107

Regression equation :   

ŷ = 12.0623 + (0.7771) x  

Predicted value of y at x =    85

ŷ = 12.0623 + (0.7771) * 85 = 78.1194  

Significance level, α =    0.05

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

95% Confidence interval :  

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

= 78.1194 - 2.3646*19.4715*√((1/9) + ((85 - 78.5556)²/(2018.2222))) = 61.411

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

= 78.1194 + 2.3646*19.4715*√((1/9) + ((85 - 78.5556)²/(2018.2222))) = 94.828