Question

1. The following table are a random sample of 8 homes, shows the relationship between kilowatt-hours (thousands) used and the number of rooms in a private single-family residences. Homes Number of KWh (thousands) a) Compute the correlation coefficient b) Can we conclude that there is a positive association betwen KWh and number of in all private single-family residences? Use th 0.05 significance level. rooms c Develop the regression equation for electricity used(Wh) based d) Determine the coefficient of determination. e) Determine the 95% prediction interval of number of KWh, in thousands, for a six- on number of rooms. room houses (Standard error of estimate -0 4873 and y--10875

Answer 1

We need to find the sums ∑x , ∑y, ∑x*y, ∑x² and ∑y²

We can use excel to find these sums :

∑x = 39 , ∑y = 33 , ∑x*y =167, ∑x² = 201 and ∑y² = 141

r = $\frac{n\sum (x*y)-\sum (x)*\sum (y)}{\sqrt{[n\sum x^2-(\sum x)^2 ]*[n\sum y^2-(\sum y)^2]}}$

r = $\frac{(8*167)-(39*33)}{\sqrt{[(8*201)-(39*39) ]*[(8*141)-(33*33)]}}$

r = 0.841209

b)

To check the significance of positive correlation we have to perform correlation test .

H₀ : The significant positive correlation exists between X and Y

H_a : The significant positive correlation does not exists between X and Y

Test statistic : r = 0.8412

Critical value : We are given  α = 0.05, and we have n = number of pairs of x and y = 8

Since we are testing for positive correlation , this is one tailed test.

Critical value = 0.621

Decision rule :

If correlation coefficient r is greater than critical value , there is significant positive correlation exists.

If correlation coefficient r is less than critical value , there is no significant positive correlation exists.

Here r = 0.8412 and critical value = 0.621

As r is greater than critical value , there is significant positive correlation exists between the X and Y

C) Regression equation :

kWh = b₀ + b₁ *number of rooms ; b₀ is Intercept and b₁ is Slope

b₁ = $\frac{n\sum (x*y)-\sum (x)*\sum (y)}{{[n\sum x^2-(\sum x)^2 ]}}$

b₁ = $\frac{(8*167)-(39*33)}{{(8*201)-(39*39)}}$

b₁ = 0.5632

b₀ = $\bar{y}-(b_1*\bar{x})$

b₀ = 4.125 - (0.5632*4.875)

b₀ = 1.3793

kWh = 1.3793 + 0.5632*number of rooms

D) Coefficient of determination (r²) = r*r = 0.8412*0.8412

r² = 0.7076

E) Prediction interval :

Lower bound = $\hat{y} - SE$

Upper bound = $\hat{y} + SE$

SE = $(t*s)*\sqrt{1+\frac{1}{n}+\frac{(x-\bar{x})^2}{SS_{xx}}}$

We are given s = 0.4873 and ( x - $\bar{x}$ )² = 10.875 , SS_xx = $\sum x^2-\frac{(\sum x)^2}{n}$ = 6.125

To find t, we can use excel function =TINV(α, d.f)  

We are given confidence level = 0.95 , therefore α = 1 - 0.95 = 0.05

d.f = n - 2 = 8 - 2 = 6

=TINV(0.05,6) = 2.4469

SE = $(t*s)*\sqrt{1+\frac{1}{n}+\frac{(x-\bar{x})^2}{SS_{xx}}}$ = $(2.4469*0.4873)*\sqrt{1+\frac{1}{8}+\frac{10.875}{6.125}}$