Question

A fast-food chain decided to carry out an experiment to assess the influence of advertising expenditure on sales. Different relative changes in advertising expenditure, compared to the previous year, were made in eight regions of the country, and resulting changes in sales levels were observed. The accompanying table shows the results

Increase in advertising expenditure (%) 0 4 14 10 9 8 6 1

Increase in sales (%) 2.4 7.2 10.3 9.1 10.2 4.1 7.6 3.5 

a. Estimate the least-squares the linear regression of the increase in sales on increase in advertising expenditure

b. Find a 90% confidence interval for the slope of the population regression line

Answer 1

We have

$a:\\X=\left[ \begin{array}{c} \begin{matrix} 1& 0\\ 1& 4\\ 1& 14\\ 1& 10\\\end{matrix}\\ \begin{matrix} 1& 9\\ 1& 8\\ 1& 6\\ 1& 1\\\end{matrix}\\\end{array} \right] \\y=\left[ \begin{array}{c} 2.4\\ 7.2\\ 10.3\\ 9.1\\ 10.2\\ 4.1\\ 7.6\\ 3.5\\\end{array} \right] \\\left[ \begin{array}{c} a\\ b\\\end{array} \right] =\left( X^TX \right) ^{-1}X^Ty=\\=\left( \left[ \begin{matrix} 1& 1& 1& 1& 1& 1& 1& 1\\ 0& 4& 14& 10& 9& 8& 6& 1\\\end{matrix} \right] \left[ \begin{array}{c} \begin{matrix} 1& 0\\ 1& 4\\ 1& 14\\ 1& 10\\\end{matrix}\\ \begin{matrix} 1& 9\\ 1& 8\\ 1& 6\\ 1& 1\\\end{matrix}\\\end{array} \right] \right) ^{-1}\left[ \begin{matrix} 1& 1& 1& 1& 1& 1& 1& 1\\ 0& 4& 14& 10& 9& 8& 6& 1\\\end{matrix} \right] \left[ \begin{array}{c} 2.4\\ 7.2\\ 10.3\\ 9.1\\ 10.2\\ 4.1\\ 7.6\\ 3.5\\\end{array} \right] =\\=\left[ \begin{matrix} 8& 52\\ 52& 494\\\end{matrix} \right] ^{-1}\left[ \begin{array}{c} 54.4\\ 437.7\\\end{array} \right] =\frac{1}{8\cdot 494-52^2}\left[ \begin{matrix} 494& -52\\ -52& 8\\\end{matrix} \right] \left[ \begin{array}{c} 54.4\\ 437.7\\\end{array} \right] =\frac{1}{1248}\left[ \begin{array}{c} 4113.2\\ 672.8\\\end{array} \right] =\\=\left[ \begin{array}{c} 3.29583\\ 0.539103\\\end{array} \right] \\y'=3.29583+0.539103x$

b:

The slope is

$\beta =0.539103$

Next,

from which

$s_{\beta}=\sqrt{\frac{\sum{\left( y_i-y_i' \right) ^2}}{\left( n-2 \right) \sum{\left( x_i-\bar{x} \right) ^2}}}=\sqrt{\frac{22.10147}{\left( 8-2 \right) \cdot 156}}=0.153664$

The confidence interval is

$\left( \beta -t_{1-\alpha /2,n-2}s_{\beta},\beta +t_{1-\alpha /2,n-2}s_{\beta} \right) =\\=\left( 0.53910-1.94318\cdot 0.15366,0.53910+1.94318\cdot 0.15366 \right) =\\=\left( 0.240511,0.837689 \right)$