Question

29. Short Answer Question We have a dataset with n = 10 pairs of observations (Li, Yi), and 72 72 T 683, 4x = 813, IM-IM: 11 n n x = 47,405, y = 56,089, 42 = 66, 731. i=1 11 i1 What is an approximate 95% confidence interval for the intercept of the line of best fit?

Answer 1

From the given data we calculate the following to get the se(b0).

=1 TS ci η

683 - 68.3 10

Σ=1 Ji 3 η

813 Ý = 81.3 10

η 1 SSxx = Σ? Ti n i=1 i=1

$SS_{XX}=47405-\frac{1}{10}\left (683 \right )^{2}$

466489 SSxx = 47405 10

$SS_{XX}=47405-46648.9=756.1$

$SS_{YY}=\sum_{i=1}^{n}y_{i}^{2}-\frac{1}{n}\left ( \sum_{i=1}^{n}y_{i} \right )^{2}$

$SS_{YY}=66731-\frac{1}{10}\left ( 813 \right )^{2}$

$SS_{YY}=66731-\frac{660969}{10}$

$SS_{YY}=66731-66096.9$

$SS_{YY}=634.1$

η SSXY - Σ - (Σ) (Σ Ji i=1

$SS_{XY}=56089-\frac{1}{10}\left ( 683\right )\left ( 813 \right )$

$SS_{XY}=56089-\frac{555279}{10}$

$SS_{XY}=56089-55527.9=561.1$

Bi SSry SS

$\hat{\beta }_{1}=\frac{561.1}{756.1}$

$\hat{\beta }_{1}=0.7421$

$\hat{\beta }_{0}=81.3-0.7421*68.3$

$\hat{\beta }_{0}=30.6146$

The sum of squares due to regression RSS=$\hat{\beta }_{1}*SS_{xy}=416.3923$

Total sum of squares TSS=$SS_{yy}=634.1$

Error SS=TSS-RSS=217.7077

Mean square error =ESS/8=27.2135.

Standard error s=5.2167

The standard error of
Bo
is given by $SE(\hat{\beta }_{0})=s*\sqrt{\frac{\sum x^2}{n*\sum x^2-(\sum x)^2}}$

  $SE(\hat{\beta }_{0})=5.2167*\sqrt{\frac{47405}{10*47405-(683)^2}}$

  $SE(\hat{\beta }_{0})=5.2167*\sqrt{\frac{47405}{474050-466489}}$

$SE(\hat{\beta }_{0})=5.2167*\sqrt{\frac{47405}{7561}}$

  $SE(\hat{\beta }_{0})=5.2167*\sqrt{6.2697}$

  $SE(\hat{\beta }_{0})=5.2167*2.5039$

$SE(\hat{\beta }_{0})=13.0623$

The 95% confidence interval for the intercept is $CI=\hat{\beta }_{0}\mp t_{8,0.95}SE(\hat{\beta }_{0})$

  $CI=30.6146\mp 2.306*13.0623$

  $CI=30.6146\mp 30.1217$

$CI=(0.4929,60.7363)$

The 95% confidence interval for the intercept is (0.4929,60.7363).