Question

A random sample of 20 workers in a factory were asked to report the age of their car and how many miles the vehicle had on it. A computer printout resulted in the following information.

Variable	Coef	SE Coef	t-ratio	Prob
Constant	7288.54	6591	1.11	<0.2826
Age	11630.6	1249	9.31	<0.0001
R sq = 82%	R sq adj = 81.1%		s = 19280

1. Find the LSRL
2. A new worker starts next week and we know that his car is 7 years old, how many miles would you expect to be on his car?
3. Interpret the slope of the LSRL in the context of this problem.
4. Find a 95% confidence interval for the slope of the LSRL
5. Without calculating it what do you know about an 85% confidence interval compared to your answer in part b?

Answer 1

The given result a simple linear model to predict age of the car.

The simple linear regression is used to predict a continuous outcome variable (y) based on one single predictor variable (x).

The mathematical formula is

y = b0 + b1*x + e

Which means:

beta1 (b1) times x, plus a constant beta0 (b0), plus an error term e

· b0 is the intercept

· b1 is the slope of the regression line

· e is the error term

From the give computer printed results,

x = Age of the car

y = miles the vehicle had on car

Intercept = Coef

The intercept (b0) is 7288.54 and coefficient for Age variable (b1), also known as the slope, is 11630

The following is the estimated regression equation:

Miles = 7288.54 + 11630 * Age of the car

If the age of the car is given then using this formula we can calculate its miles.

Question1: Find the LSRL

Answer: A regression line (LSRL - Least Squares Regression Line) is a straight line that describes how a response variable y changes as an explanatory variable x changes.

To find LSRL:

Question2: A new worker starts next week and we know that his car is 7 years old, how many miles would you expect to be on his car?

Answer: Miles = 7288.54 + (11630 * Age of the car)

           Miles = 7288.54 + 11630*7 = 88698.54

miles would you expect to be on his car is 88698.54

Question3: Interpret the slope of the LSRL in the context of this problem.

Answer:

For a car with age7 we can expect an increase of 81410 (i.e., 11630 *7) miles

That is,    Miles = 7288.54 + 11630*7 = 88698.54

Question4: Find a 95% confidence interval for the slope of the LSRL

Answer:

Slope b1 = 11630

Confidence level = 95%

Compute alpha (α):

α = 1 - (confidence level / 100) = 1 - (95/100) = 0.05

Find the critical probability (p):

P = 1 - α/2 = 1 - 0.05/2 = 0.975

Find the degrees of freedom (df):

Here n = 20 (sample of 20 workers)

df = n - 2 = 20 - 2 = 18.

Find critical value:

From the T distribution critical value table we can identify critical value for degrees of freedom = 18 and α = 0.05

critical value = 1.734

Compute margin of error (ME):

    ME = critical value * standard error

SE Coef in the model represents the standard error.

    ME = 1.734 * 1249 = 2165.766

The range of the confidence interval is defined by the sample statistic + or - margin of error

Therefore, the 95% confidence interval for this sample is 11630 + or - 2165.766

85% confidence interval

Confidence level = 85%

α = 1 - (confidence level / 100) = 1 - (85/100) = 0.15

Degrees of freedom (df) = n-2 = 20 – 2 = 18

Critical value = 1.504

    ME = critical value * standard error = 1.504 * 1249 = 1878.496

Therefore, the 85% confidence interval for this sample is 11630 + or - 1878.496