Question

Software millionaires and birthdays. Refer to Exercise 11.23 (p. 655) and the study of software millionaires and their birthdays. The data are reproduced on p. 663.

a. Find SSE ${\mathit{s}}^{\mathbf{2}}$and s for the simple linear regression model relating the number (y) of software millionaire birthdays in a decade to the total number (x) of U.S. births.

b. Find SSE ${\mathit{s}}^{\mathbf{2}}$and s for the simple linear regression model relating the number (y) of software millionaire birthdays in a decade to the number (x) of CEO birthdays.

c. Which of the two models' fit will have smaller errors of prediction? Why?

Decade	Total U.S. Births (millions)	Number of Software Millionaire Birthdays	Number of CEO Birthdays (in a random sample of 70 companies from the Fortune 500 list)
1920	28.582	3	2
1930	24.374	1	2
1940	31.666	10	23
1950	40.530	14	38
1960	38.808	7	9
1970	33.309	4	0

Short answer

1. SSE is 34051.540643, ${\mathit{s}}^{\mathbf{2}}$ 8512.88516075, and s is 92.2653.
2. SSE is 2296.75, ${\mathit{s}}^{\mathbf{2}}$ 1148.375, and s is 33.8877.
3. Model 2.

Step-by-step solution

Introduction

The standard error of the mean is the most commonly reported form of standard error (SE or SEM). The standard error may also be used to find other statistics such as medians and proportions. The standard error is a measure that assesses the difference between a population parameter and a sample statistic.

Calculate SSE s2 and s via the number (y) of software millionaire birthdays in a decade to the number (x) of U.S Births

\(\begin{aligned}S{S_{xy}} &= \sum {{x_i}} {y_i} - \frac{{\sum {{x_i}} \sum {{y_i}} }}{n}\\ &= 139.092 - \frac{{197.269 x 78}}{6}\\ &= 139.092 - \frac{{15,386.982}}{6}\\ &= \frac{{139.092 - 15,386.982}}{6}\end{aligned}\)

\(\begin{aligned} &= - \frac{{15,247.89}}{6}\\ &= 2,541.315\end{aligned}\)

\(\begin{aligned}S{S_{xx}} &= {\sum {{x_i}} ^2} - \frac{{{{\left( {\sum {{x_i}} } \right)}^2}}}{n}\\ &= 6671.989401 - \frac{{{{\left( {197.269} \right)}^2}}}{6}\\ &= 6671.989401 - \frac{{38915.058361}}{6}\\ &= \frac{{40,031.936406 - 38915.058361}}{6}\end{aligned}\)

\(\begin{aligned} &= \frac{{1116.878045}}{6}\\ &= 186.1463\end{aligned}\)

\(\begin{aligned}\widehat {{\beta _1}} &= \frac{{S{S_{xy}}}}{{S{S_{xx}}}}\\ &= \frac{{2541.315}}{{186.1463}}\\ &= 13.6522\end{aligned}\)

\(\begin{aligned}S{S_{yy}} &= \sum {y^2} - \frac{{{{\left( {\sum y} \right)}^2}}}{n}\\ &= 371 - \frac{{{{\left( {78} \right)}^2}}}{6}\\ &= 371 - \frac{{6084}}{{12}}\\ &= \frac{{2226 - 6084}}{6}\end{aligned}\)

\(\begin{aligned} &= - \frac{{3858}}{6}\\ &= - 643\end{aligned}\)

\(\begin{aligned}SSE &= S{S_{yy}} - \widehat {{\beta _1}}S{S_{xy}}\\ &= 643 - (13.6522)(2541.315)\\ &= 643 - 34694.540643\\ &= 34051.540643\end{aligned}\)

\(\begin{aligned} {s^2} &= \frac{{SSE}}{{n - 2}}\\ &= \frac{{34051.540643}}{{6 - 2}}\\ &= \frac{{34051.540643}}{4}\\ &= 8512.88516075\end{aligned}\)

\(\begin{aligned}s &= \sqrt {{s^2}} \\ &= \sqrt {8512.88516075} \\ &= 92.2653\end{aligned}\)

Therefore, the value of SSE is 34051.540643, ${\mathit{s}}^{\mathbf{2}}$ 8512.88516075, and s is 92.2653.

Calculate SSE s2  and s via the number (y) of software millionaire birthdays in a decade to the number (x) of CEO birthdays

\(\begin{aligned}S{S_{xy}} &= \sum {{x_i}} {y_i} - \frac{{\sum {{x_i}} \sum {{y_i}} }}{n}\\ &= 1646 - \frac{{74 x 39}}{6}\\ &= 1646 - \frac{{2886}}{6}\\ &= \frac{{9876 - 2886}}{6}\end{aligned}\)

\(\begin{aligned} &= \frac{{6990}}{6}\\ &= 1665\end{aligned}\)

\(\begin{aligned}S{S_{xx}} &= {\sum {{x_i}} ^2} - \frac{{{{\left( {\sum {{x_i}} } \right)}^2}}}{n}\\ &= 2062 - \frac{{{{\left( {74} \right)}^2}}}{6}\\ &= 2062 - \frac{{5476}}{6}\\ &= \frac{{12372 - 5476}}{6}\end{aligned}\)

\(\begin{aligned} &= \frac{{6896}}{6}\\ &= 1149.4\end{aligned}\)

\(\begin{aligned}\widehat {{\beta _1}} &= \frac{{S{S_{xy}}}}{{S{S_{xx}}}}\\ &= \frac{{1665}}{{1149.4}}\\ &= 1.45\end{aligned}\)

\(\begin{aligned}S{S_{yy}} &= \sum {y^2} - \frac{{{{\left( {\sum y} \right)}^2}}}{n}\\ &= 371 - \frac{{{{\left( {39} \right)}^2}}}{6}\\ &= 371 - \frac{{1521}}{6}\\ &= \frac{{2226 - 1521}}{6}\end{aligned}\)

\(\begin{aligned} &= \frac{{705}}{6}\\ &= 117.5\end{aligned}\)

\(\begin{aligned}SSE &= S{S_{yy}} - \widehat {{\beta _1}}S{S_{xy}}\\ &= 117.5 - (1.45)(1665)\\ &= 117.5 - 2414.25\\ &= 2296.75\end{aligned}\)

\(\begin{aligned} {s^2} &= \frac{{SSE}}{{n - 2}}\\ &= \frac{{2296.75}}{{4 - 2}}\\ &= \frac{{2296.75}}{2}\\ &= 1148.375\end{aligned}\)

\(\begin{aligned}s &= \sqrt {{s^2}} \\ &= \sqrt {1148.375} \\ &= 33.8877\end{aligned}\)

Therefore, the value of SSE is 2296.75, ${\mathit{s}}^{\mathbf{2}}$ 1148.375,and s is 33.8877.

The two models' fits will have smaller prediction errors

From steps 2 and 3, it is clear that Model 2 fits, as it has a smaller prediction error.