Question

Confidence Intervals for a Regression Coefficients A confidence interval for the regression coefficient b1 is expressed

\(\begin{array}{l}{b_1} - E < {\beta _1} < {b_1} + E\\\end{array}\)

Where

\(E = {t_{\frac{\alpha }{2}}}{s_{{b_1}}}\)

The critical t score is found using n –(k+1) degrees of freedom, where k, n, and sb1 are described in Exercise 17. Using the sample data from Example 1, n = 153 and k = 2, so df = 150 and the critical t scores are \( \pm \)1.976 for a 95% confidence level. Use the sample data for Example 1, the Stat diskdisplay in Example 1 on page 513, and the Stat Crunchdisplay in Exercise 17 to construct 95% confidence interval estimates of \({\beta _1}\) (the coefficient for the variable representing height) and\({\beta _2}\) (the coefficient for the variable representing waist circumference). Does either confidence interval include 0, suggesting that the variable be eliminated from the regression equation?

Short answer

The 95% confidence interval of\({\beta _1}\)is (0.6288,0.9098).

The 95% confidence interval of\({\beta _2}\)is equal to (0.9427,1.0763).

Both the confidence intervals do not include 0. Hence, the variables need not be eliminated from the regression equation as they are significant in predicting the variable “Weight”.

Step-by-step solution

Given information

The 95% confidence intervals for the regression coefficients \({\beta _1}\) and \({\beta _0}\)need to be constructed. The sample size (n) is equal to 153. The number of predictors (k) is equal to 2. The degrees of freedom are computed using the formula n-(k+1). The critical value of t is given to be equal to 1976.

Calculation of the margins of error

Margin of error corresponding to\({\beta _1}\):

Let\({b_1}\)be the estimated value of\({\beta _1}\).

Using the output from exercise 17, the value of standard error of\({b_1}\)is equal to :\({s_{{b_1}}} = 0.0711\). The value of\({b_1}\)is equal to 0.7693.

It is given that\({t_{\frac{\alpha }{2}}} = 1.976\).

Substituting the required values in the formula of E, the following of E is obtained:

\(\begin{array}{c}{E_1} = {t_{\frac{\alpha }{2}}}{s_{{b_1}}}\\ = 1.976 \times 0.0711\\ = 0.1405\end{array}\)

Therefore, the margin of error is 0.1405.

Margin of error corresponding to\({\beta _1}\):

Let\({b_2}\)be the estimated value of\({\beta _2}\).

Using the output from exercise 17, the value of standard error of\({b_2}\)is equal to\({s_{{b_2}}} = 0.0338\). The value of\({b_2}\)is equal to 1.0095.

It is given that\({t_{\frac{\alpha }{2}}} = 1.976\).

Substituting the required values in the formula of E, the following of E is obtained:

\(\begin{array}{c}{E_2} = {t_{\frac{\alpha }{2}}}{s_{{b_1}}}\\ = 1.976 \times 0.0338\\ = 0.0668\end{array}\)

Therefore, the margin of error is 0.0668.

Step 3:Computation of the confidence intervals

Confidence interval for\({\beta _1}\):

The 95% confidence interval for\({\beta _1}\)is computed as follows:

\(\begin{array}{c}{b_1} - {E_1} < {\beta _1} < {b_1} + {E_1}\\0.7693 - 0.1405 < {\beta _1} < 0.7693 + 0.1405\\0.6288 < {\beta _1} < 0.9098\end{array}\)

Therefore, the 95% confidence interval of\({\beta _1}\)is (0.6288,0.9098).

Confidence interval for\({\beta _2}\):

The 95% confidence interval for\({\beta _2}\)is computed as follows:

\(\begin{array}{c}{b_2} - {E_2} < {\beta _2} < {b_2} + {E_2}\\1.0095 - 0.0668 < {\beta _2} < 1.0095 + 0.0668\\0.9427 < {\beta _2} < 1.0763\end{array}\)

Therefore, the confidence interval of \({\beta _2}\) is equal to (0.9427,1.0763).

Step 4:Interpretation of the confidence intervals

Let\({\beta _1}\)be the regression coefficient of the predictor “Height”.

The confidence interval of\({\beta _1}\)is equal to (0.6288, 0.9098).

Since the confidence interval does not contain the value 0, the variable “Height” is significant in predicting the variable “Weight”.

Let\({\beta _2}\)be the regression coefficient of the predictor “Waist”.

The confidence interval of\({\beta _1}\)is equal to (0.9427, 1.0763).

Since the confidence interval does not contain the value 0, the variable “Waist” is significant in predicting the variable “Weight”.