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I only need help with E, F and J please! Statistics Assignment 15 1. A local...

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 pe
e) Find the standard error of estimate S Find a 90% prediction interval for the number of cars sold for a week when 12 one mi
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
e) Find the standard error of estimate S Find a 90% prediction interval for the number of cars sold for a week when 12 one minute ads were run. g) Looking at the scatter diagram do you think the correlation coefficient will be positive, negative, or zero? h) Find r, the correlation coefficient then find the coefficient of determination. Test the claim that the population correlation coefficient is positive use a 5% level of significance. j) Test the claim that the slope 3 of the population least-squares line is positive at the 5% level of significance.
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Answer #1
X Y XY
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.  

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