Question

In an effort to determine the cost of air conditioning, a resident in College Station,TX, recorded daily values of the variables

Tavg = mean temperature

Kwh = electricity consumption

for the period from September 19 through November 4.

tavg kwh tavg kwh tavg kwh tavg kwh

77.5 45 77.5 55 68 50 75 41

80 73 79 57 66.5 37 75.5 51

78 43 80 68 69 43 71.5 34

78.5 61 79 73 70.5 42 63 19

77.5 52 76 57 63 25 60 19

83 56 76 51 64 31 64 30

83.5 70 75.5 55 64.5 31 62.5 23

81.5 69 79.5 56 65 32 63.5 35

75.5 53 78.5 72 66.5 35 73.5 29

69.5 51 82 73 67 32 68 55

70 39 71.5 69 66.5 34 77.5 56

73.5 55 70 38 67.5 35

(a) Give and interpret a 98% confidence interval for ?.

(b) Give and interpret a 90% confidence interval for ?1.

(c) Give and interpret a 90% interval for the average electrify consumption for days that are 79 degrees Fahrenheit.

(d) Tomorrow it is supposed to be an average of 67 degrees Fahrenheit in College Station, TX. Predict tomorrow's electricity consumption using a 90% interval.

(e) Run an appropriate hypothesis test to determine if there is statistically significant evidence that the average electricity consumption changes with the mean daily temperature.

Answer 1

Regression

Regression Equation

Kwh = -97.92 + 2 * tavg

Y = α + β₁ X

a)

At alpha = 0.02,

df = 47-1 = 46

t_Critical = 2.404

Hence, 98% CI for α = -97.22 +/- 2.404 * 13.86 = {-131.25, -64.6}

Interpretation

There is 98% Probability that the true mean value of α lies in the interval {-131.25, -64.6}

b)

At alpha = 0.1,

df = 47-1 = 46

t_Critical = 1.676

Hence, 90% CI for β₁ = 2 +/- 1.676 * 0.19 = {1.68, 2.32}

Interpretation

There is 90% Probability that the true mean value of β₁ lies in the interval {1.68, 2.32}

c)

When, tavg = 79,

Using the Regression Equation

Kwh = -97.92 + 2 * 79 = 60.16

At alpha = 0.1,

df = 47-1 = 46

t_Critical = 1.676

Hence, 90% CI for Kwh = 60.16 +/- 1.676 * 8.39 = {46.1, 74.21}

Interpretation

There is 90% Probability that the true mean value of Kwh lies in the interval {46.1, 74.21}

d)

When, tavg = 67,

Using the Regression Equation

Kwh = -97.92 + 2 * 67 = 36.14

At alpha = 0.1,

df = 47-1 = 46

t_Critical = 1.676

Hence, 90% CI for Kwh = 36.14 +/- 1.676 * 8.39 = {22.09, 50.20}

e)

From the Regression Output, r = 0.8426

alpha = 0.05

Null and Alternate Hypothesis

H₀: Sigma = 0

H_a: Sigma <> 0

Test Statistic

t = r * (n-1)^1/2 / (1-r²)^1/2 = 10.61

P-value = TDIST(10.61,46,2) = 0.00000000000006

Since, the p-value is less than 0.05, we reject the null hypothesis ie there is significant correlation between the two variables ie electricity consumption changes with the mean daily temperature.