Chapter 11.2, Problem 29E

a.

To determine

To explain: The null and alternative hypotheses for a significance test.

Explanation of Solution

Given:

The observed value ( O_i ) is:

Goal	FEMALE	MALE
HSC-HM	14	31
HSC-LM	7	18
LSC-HM	21	5
LSC-LM	25	13

Calculation:

To test whether there is a difference of sports goals for male and female undergraduates at the university or not.

The null and alternative hypotheses are:

  $\begin{array}{l}{H}_0:\left(\begin{array}{l}\text{There is no difference in the distribution sports goals }\\ \text{for male and female undergraduates at the university}\end{array}\right)\\ H_1:\left(\begin{array}{l}\text{There is a difference in the distribution sports goals }\\ \text{for male and female undergraduates at the university}\end{array}\right)\end{array}$

b.

To determine

To find: The expected counts for the given data set.

Explanation of Solution

Formula used:

The expected value ( E_i ), can be calculated as:

  $E_i{}_j=\frac{\text{Row sum}\left(i\right)\times \text{Column Sum}\left(j\right)}{\text{Total sum}}$

Calculation:

From the given table,

Goal	FEMALE	MALE	Total
HSC-HM	14	31	45
HSC-LM	7	18	25
LSC-HM	21	5	26
LSC-LM	25	13	38
Total	67	67	134

The expected value ( E_i ), can be calculated as:

Goal	FEMALE	MALE
HSC-HM	$\frac{67\times 45}{134}=22.50$	$\frac{67\times 31}{134}=15.50$
HSC-LM	$\frac{67\times 25}{134}=12.50$	$\frac{67\times 25}{134}=12.50$
LSC-HM	$\frac{67\times 26}{134}=13$	$\frac{67\times 26}{134}=13$
LSC-LM	$\frac{67\times 38}{134}=19$	$\frac{14\times 31}{134}=22.50$

c.

To determine

To find: The chi-square statistic.

Answer to Problem 29E

The chi-square test value, is 24.8978.

Explanation of Solution

Formula used:

The chi-square test statistics value:

  ${\chi }^2{}_{statistic}={\displaystyle \sum \frac{{\left(O_i-E_i\right)}^2}{E_i}}$

Calculation:

The chi-square statistic, can be calculated as:

  $\begin{array}{c}{\chi }^2{}_{statistic}=\left(\begin{array}{l}\frac{{\left(14-22.50\right)}^2}{22.50}+\frac{{\left(31-15.50\right)}^2}{15.50}+\frac{{\left(7-12.50\right)}^2}{12.50}+\frac{{\left(18-12.50\right)}^2}{22.50}\\ +\frac{{\left(21-13\right)}^2}{13}+\frac{{\left(5-13\right)}^2}{13}+\frac{{\left(25-19\right)}^2}{19}+\frac{{\left(13-22.50\right)}^2}{22.50}\end{array}\right)\\ =24.8978\end{array}$

Thus, the chi-square test value, is 24.8978.