Question

To predict a bear’s weight, data was taken from 54 anesthetized bears on their age, gender (M=1, F=2), head width_, neck size,  overall length, and chest size.

ANOVA
	df	SS
Regression	6	741937.3181
Residual	47	44346.0152
Total	53	786283.3333

	Coefficients	Standard Error
Intercept	-209.575	39.453
AGE	0.526	0.225
GENDER	-12.186	11.503
HEADWTH	-0.413	5.041
NECK	3.811	2.607
LENGTH	-0.463	0.914
CHEST	9.459	1.443

Use above data to construct a 95% confidence interval for the variable ' AGE' and answer following:

Name the critical value needed to compute the error

Compute the Margin of Error (3 decimal places)  

Name the lower limit of the confidence interval (3 decimal places)

Name the upper limit of the confidence interval (3 decimal places)

What percent of variation(1 decimal place) in a bear's weight is NOT explained by the combination of variables in the regression equation?

Answer 1

Critical value:

The total number of observations (n) is given by:

The confidence level (c) is 0.95, so the significance level is 1- 0.95 = 0.05.

At the significance level 0.05 and degrees of freedom 52 ( = n – 2), the two tailed critical value obtained from the t-table is +/- 2.007.

Therefore, the critical value is +/- 2.007.

Construct 95% confidence interval:

On the basis of given output, the coefficient (b) and standard error corresponding to variable age are 0.526 and 0.225 respectively.

The 95% confidence interval for the variable Age is given by,

Therefore, the required confidence interval is (0.074, 0.978).

Margin of error (ME):

Therefore, the required margin of error is 0.452.

The lower limit of the confidence interval is 0.074.

The upper limit of the confidence interval is 0.978.

The unexplained variation in the regression equation can be calculated as:

Therefore, 5.6% of variation in a bear’s weight is not explained by the combination of variables in the regression equation.