Statistics - Regression Intercept Confidence Interval

Formula

${R = \beta_0 \pm t(1 - \frac{\alpha}{2}, n-k-1) \times SE_{\beta_0} }$

Where −

${\beta_0}$ = Regression intercept.
${k}$ = Number of Predictors.
${n}$ = sample size.
${SE_{\beta_0}}$ = Standard Error.
${\alpha}$ = Percentage of Confidence Interval.
${t}$ = t-value.

Example

Problem Statement:

Compute the Regression Intercept Confidence Interval of following data. Total number of predictors (k) are 1, regression intercept ${\beta_0}$ as 5, sample size (n) as 10 and standard error ${SE_{\beta_0}}$ as 0.15.

Solution:

Let us consider the case of 99% Confidence Interval.

Step 1: Compute t-value where ${ \alpha = 0.99}$ .

${ = t(1 - \frac{\alpha}{2}, n-k-1) \\[7pt] = t(1 - \frac{0.99}{2}, 10-1-1) \\[7pt] = t(0.005,8) \\[7pt] = 3.3554 }$

Step 2: ${\ge}$ Regression intercept:

${ = \beta_0 + t(1 - \frac{\alpha}{2}, n-k-1) \times SE_{\beta_0} \\[7pt] = 5 - (3.3554 \times 0.15) \\[7pt] = 5 - 0.50331 \\[7pt] = 4.49669 }$

Step 3: ${\le}$ Regression intercept:

${ = \beta_0 - t(1 - \frac{\alpha}{2}, n-k-1) \times SE_{\beta_0} \\[7pt] = 5 + (3.3554 \times 0.15) \\[7pt] = 5 + 0.50331 \\[7pt] = 5.50331 }$

As a result, Regression Intercept Confidence Interval is ${4.49669}$ or ${5.50331}$ for 99% Confidence Interval.

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