Question

25. Short Answer Question We have a dataset with n 10 pairs of observations (1, y.), and di = 683, Σ» - και Yi = 813, EM-IM n ni 27 = 47, 405, * Vi = 56,089, Lu= 66, 731. What is an approximate 99% confidence interval for the slope of the line of best fit?

Answer 1

Solution:

Given:

Σ Τ, = 683, 9: 813, 1 ή Σ' = 47, 405, Σαμι = 56, 089. Σ = 66, 731. i1

Confidence interval formula for slope of line of regression.

(61 - E bi + E

where

E = tc X SE

Se SE 1 SSII

$Se=\sqrt{\frac{SSE}{n-2}}$

SS2 SSE = SSyy SS TT

$b_{1}=\frac{SS_{xy}}{SS_{xx}}$

SSpy – Στy – (Στ και Συ / η

$SS_{xy}=56089 \: \: -\: \: \left ( 683 \times 813 \: \: /\: \: 10\: \: \right )$

$SS_{xx}=\sum x^{2} \: \: -\: \: \left ( \sum x\times \sum x \: \: /\: \: n\: \: \right )$

$SS_{xx}=47405 \: \: -\: \: \left ( 683 \times 683 \: \: /\: \: 10\: \: \right )$

$SS_{yy}=\sum y^{2} \: \: -\: \: \left ( \sum y\times \sum y \: \: /\: \: n\: \: \right )$

$SS_{yy}= 66731 \: \: -\: \: \left ( 813 \times 813 \: \: /\: \: 10\: \: \right )$

Thus

$b_{1}=\frac{SS_{xy}}{SS_{xx}}$

$b_{1}=\frac{561.1 }{756.1 }$

$\mathbf{{\color{DarkOrange} b_{1}=0.7421}}$

SS2 SSE = SSyy SS TT

$SSE = 634.1 -\frac{561.1 ^{2}}{756.1 }$

thus

$Se=\sqrt{\frac{SSE}{n-2}}$

$Se=\sqrt{\frac{217.7090 }{10-2}}$

$Se=\sqrt{\frac{217.7090 }{8}}$

$Se=\sqrt{ 27.2136 }$

thus

$SE_{b1}=\frac{S_{e}}{\sqrt{SS_{xx}}}$

$SE_{b1}=\frac{5.2167 }{\sqrt{756.1 }}$

$SE_{b1}=\frac{5.2167 }{ 27.49727 }$

and

t_c is t critical value for c = 99%  confidence level

Thus two tail area = 1 - c = 1 - 0.99 = 0.01

df = n - 2 =  10 - 2 = 8

Look in  t table for df = 8 and two tail area = 0.01 and find t critical value

t5 t 55 t 875 t Table cum. prob one-tail two-tails df 0.50 1.00 0.15 t.80 0.20 0.40 0.25 0.50 t86 0.05 0.10 0.10 0.20 0.30 0.025 0.05 0.01 0.02 0.005 0.01 000 OWN 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 1.000 0.816 0.765 0.741 0.727 0.718 0.711 0.796 1.376 1.061 0.978 0.941 0.920 0.906 0.896 0.8.89 1.963 1.386 1.250 1.190 1.156 1.134 1.119 1.108 3.078 1.886 1.638 1.533 1.476 1.440 1.415 1.397 6.314 2.920 2.353 2.132 2.015 1.943 1.895 1.860 12.71 4.303 3.182 2.776 2.571 2.447 2.365 2.306 31.82 6.965 4.541 3.747 3.365 3.143 2.998 2.896 63.66 9.925 5.841 4.804 4.032 3.707 3.499 3.355

t_c = 3.355

thus

E = tc X SE

$E = 3.355 \times 0.189716$

$\mathbf{{\color{DarkOrange} E = 0.636497}}$

thus

(61 - E bi + E

$\mathbf{{\color{DarkGreen} (0.1056 \: \: ,\: \: 1.3786)}}$

_{(Round final answer to specified number of decimal places)}