Question

utility company would like to predict the monthly heating bill for a household in a certain county in January. A random sample of households in the county was selected and their January heating bill was recorded along with the variables SF, Age, and Temp, with the results shown i i Click the icon to view the regression output. the accompanying regression output. Complete parts a and b below. a. Test the significance of each independent variable using a 0.01. For b, and B. let j- 1 represent the variable SF, let j 2 represent the variable Age, and let j-3 represent the variable Temp. Test the significance of the variable SF. Determine the null and alternative hypotheses. Regression Output H: B SF: the square footage of the house Age: the age of the current heating system, in years Temp: the thermostat setting, in degrees Fahrenheit, during the day ANOVA SS 102092.2343 di Regression Residual Total 28 61515.7549 31 163607.9892 Standard Error Coefficients Intercept 653.4885 322.4168 Enter your answer the edit fields and then click Check Answer. Age 20438 5064 10.8667 5.0054 Temp parts remaining Swer

What is test statistic?

What is p-value?

b. Construct a 95% confidence interval for each regression coefficient and interpret its meaning.

Answer 1

a.

H0: $\beta_1 = 0$

H1: $\beta_1 \ne 0$

Test Statistic = Coeff / Standard error = 0.0935 / 0.0207 = 4.517

Degree of freedom , df = df for Residual = 28

P-value = 2 * P(t > 4.517) = 0.0001

Since p-value is less than 0.05 significance level, we reject null hypothesis H0 and conclude that there is significant evidence that $\beta_1 \ne 0$ and variable SF is significant.

H0:

52=0

H1:

B20

Test Statistic = Coeff / Standard error = 2.0438 / 1.5064 = 1.357

Degree of freedom , df = df for Residual = 28

P-value = 2 * P(t > 1.357) = 0.1856

Since p-value is greater than 0.05 significance level, we fail to reject null hypothesis H0 and conclude that there is no significant evidence that and variable Age is significant.

B20

H0:

B3=0

H1:

B3 0

Test Statistic = Coeff / Standard error = 10.8667 / 5.0054 = 2.171

Degree of freedom , df = df for Residual = 28

P-value = 2 * P(t > 2.171) = 0.03856

Since p-value is less than 0.05 significance level, we reject null hypothesis H0 and conclude that there is significant evidence that and variable Temp is significant.

B3 0

(b)

Critical value of t at 95% confidence interval and df = 28 is 2.048

For ,

31

Margin of error = t * Standard error = 2.048 * 0.0207 = 0.0424

95% confidence interval is,

(0.0935 - 0.0424, 0.0935 + 0.0424)

(0.0511, 0.1359)

We're 95% confident that the interval (0.0511, 0.1359) captured the true value of .

31

For ,

2

Margin of error = t * Standard error = 2.048 * 1.5064 = 3.0851

95% confidence interval is,

(2.0438 - 3.0851, 2.0438 + 3.0851)

(-1.0413, 5.1289)

We're 95% confident that the interval (-1.0413, 5.1289) captured the true value of ​​​​​​​.

2

For $\beta_3$,

Margin of error = t * Standard error = 2.048 * 5.0054 = 10.2511

95% confidence interval is,

(10.8667 - 10.2511, 10.8667 + 10.2511)

(0.6156, 21.1178)

We're 95% confident that the interval (0.6156, 21.1178) captured the true value of $\beta_3$.