Question

I only need help with E, F and J please!

Statistics Assignment 15 1. A local car dealership has been running 1-minute ads on the local TV station. During a 10-week period, the dealer kept a weekly record of the number of TV ads versus the number y of cars sold. The results are given below: y 6 15 20 31 0 10 14 16 25 28 16 20 28 40 18 25 10 12 8 15

Answer 1

X	Y	XY	X²	Y²
6	15	90	36	225
20	31	620	400	961
0	10	0	0	100
14	16	224	196	256
25	28	700	625	784
16	20	320	256	400
28	40	1120	784	1600
18	25	450	324	625
10	12	120	100	144
8	15	120	64	225
Ʃx =	Ʃy =	Ʃxy =	Ʃx² =	Ʃy² =
145	212	3764	2785	5320

Sample size, n =	10
x̅ = Ʃx/n = 145/10 =	14.5
y̅ = Ʃy/n = 212/10 =	21.2
SSxx = Ʃx² - (Ʃx)²/n = 2785 - (145)²/10 =	682.5
SSyy = Ʃy² - (Ʃy)²/n = 5320 - (212)²/10 =	825.6
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 3764 - (145)(212)/10 =	690

e)

Sum of Square error, SSE = SSyy -SSxy²/SSxx = 825.6 - (690)²/682.5 =    128.0175824

Standard error, se = √(SSE/(n-2)) = √(128.01758/(10-2)) =    4.00027

f)

Slope, b = SSxy/SSxx = 690/682.5 =    1.010989

y-intercept, a = y̅ -b* x̅ = 21.2 - (1.01099)*14.5 =    6.540659

Regression equation :   

ŷ = 6.5407 + (1.011) x  

Predicted value of y at x =    12

ŷ = 6.5407 + (1.011) * 12 = 18.6725  

Critical value, t_c = T.INV.2T(0.1, 8) = 1.8595  

90% Prediction interval :  

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

= 18.6725 - 1.8595*4.0003*√(1 + (1/10) + ((12 - 14.5)²/(682.5))) =    10.8383

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

= 18.6725 + 1.8595*4.0003*√(1 + (1/10) + ((12 - 14.5)²/(682.5))) =    26.5067

j)

Null and alternative hypothesis:  

Ho: β₁ = 0  

Ha: β₁ > 0  

Test statistic:  

t = b/(se/√SSxx) =    6.6025

df = n-2 =    8

p-value = T.DIST.2T(ABS(6.6025), 8) =    0.0001

Conclusion:  

p-value < α, Reject the null hypothesis.