Chapter 10.7, Problem 18E

To determine

To find:

To show that there is a relationship between age( by college classification) and destination.

Answer to Problem 18E

Solution:

There is no relationship between age( by college classification) and destination.

Explanation of Solution

The following table represents the random sample of college Spring Break vacationers produces the results is,

	Observed Sample of 192 College Students
	Beach	Mountains	City	Home	Total
Freshman	19	2	7	24	52
Sophomore	15	4	3	20	42
Junior	18	1	9	19	47
Senior	21	6	4	20	51
Total	73	13	23	83	192

Step 1: To state the null and alternative hypotheses.

Let the null hypothesis be that relationship between age( by college classification) and destination are independent of one another.

Null hypothesis: $H_0$: Age( by college classification) and destination are independent.

Alternative hypothesis: $H_a$: Age( by college classification) and destination are dependent.

Step 2: To determine which distribution to use for the test statistic, and state the level of significance.

To determine there is a relationship between age( by college classification) and destination. Since we assume that the necessary condition that have been met. So, use the chi-square test for this association and the level of siginificance is $\alpha =0.05$.

Step 3: To calculate the test statistic.

First calculate the expected value for each cell in the contingency table.

Formula for calculating expected value of a frequency in a contingency table:

The expected value of the frequency for the i^th possible outcome in a contingency table is given by

$E_i=\frac{\left(rowtotal\right)\left(columntotal\right)}{n}$

Where n is the sample size.

From the above table represent the value of n is 192.

The expected values of for age( by college classification) and Beach is given below,

$\begin{array}{rll}{E}_{beachandfreshman}& =& \frac{73\times 52}{192}\\ & =& \frac{3796}{192}\\ & =& 19.7708\\ E_{beachand\text{sophomore}}& =& \frac{73\times 42}{192}\\ & =& \frac{3066}{192}\\ & =& 15.9688\\ E_{beachandjunior}& =& \frac{73\times 47}{192}\\ & =& \frac{3431}{192}\\ & =& 17.8698\\ E_{beachandsenior}& =& \frac{73\times 51}{192}\\ & =& \frac{3723}{192}\\ & =& 19.3906\end{array}$

The expected values of for age( by college classification) and mountains is given below,

$\begin{array}{rll}{E}_{mountainandfreshman}& =& \frac{13\times 52}{192}\\ & =& \frac{676}{192}\\ & =& 3.5208\\ E_{montainand\text{sophomore}}& =& \frac{13\times 42}{192}\\ & =& \frac{546}{192}\\ & =& 2.8438\end{array}$

$\begin{array}{rll}{E}_{mountainandjunior}& =& \frac{13\times 47}{192}\\ & =& \frac{611}{192}\\ & =& 3.1823\\ E_{mountainandsenior}& =& \frac{13\times 51}{192}\\ & =& \frac{663}{192}\\ & =& 3.4531\end{array}$

The expected values of for age( by college classification) and city is given below,

$\begin{array}{rll}{E}_{cityandfreshman}& =& \frac{23\times 52}{192}\\ & =& \frac{1196}{192}\\ & =& 6.2292\\ E_{cityand\text{sophomore}}& =& \frac{23\times 42}{192}\\ & =& \frac{966}{192}\\ & =& 5.0313\\ E_{cityandjunior}& =& \frac{23\times 47}{192}\\ & =& \frac{1081}{192}\\ & =& 5.6302\\ E_{cityandsenior}& =& \frac{23\times 51}{192}\\ & =& \frac{1173}{192}\\ & =& 6.1094\end{array}$

The expected values of for age( by college classification) and home is given below,

$\begin{array}{rll}{E}_{\mathrm{hom}eandfreshman}& =& \frac{83\times 52}{192}\\ & =& \frac{3796}{192}\\ & =& 22.4792\\ E_{\mathrm{hom}eand\text{sophomore}}& =& \frac{83\times 42}{192}\\ & =& \frac{3486}{192}\\ & =& 18.1563\end{array}$

$\begin{array}{rll}{E}_{\mathrm{hom}eandjunior}& =& \frac{83\times 47}{192}\\ & =& \frac{3901}{192}\\ & =& 20.3177\\ E_{\mathrm{hom}eandsenior}& =& \frac{83\times 51}{192}\\ & =& \frac{4233}{192}\\ & =& 22.0469\end{array}$

The table represents the contingency table of expected values by using the above formula is given below,

	Expected values
	Beach	Mountains	City	Home	Total
Freshman	19.7708	3.5208	6.2292	22.4792	52
Sophomore	15.9688	2.8438	5.0313	18.1563	42
Junior	17.8698	3.1823	5.6302	20.3177	47
Senior	19.3906	3.4531	6.1094	22.0469	51
Total	73	13	23	83	192

Formula for calculating test statistic for a Chi-Square test for association:

The test statistic for a chi-square test for association is given below,

${\chi }^2={\displaystyle \sum _{i=1}^n\frac{{\left(O_i-E_i\right)}^2}{E_i}}$

Where O_i, is the observed frequency for the i^th possible outcome and E_i is the expected frequency for the i^th possible outcome and n is the sample size.

From the above table represent the value of n is 192.

The test statistic, ${\chi }^2$, for a chi- square test for association using the above contingency tables is given below,

$\begin{array}{rll}{\chi }^2& =& {\displaystyle \sum _{i=1}^n\frac{{\left(O_i-E_i\right)}^2}{E_i}}\\ & =& \{\begin{array}{l}\frac{{\left(O_{freshman}-E_{beach}\right)}^2}{E_{beach}}+\frac{{\left(O_{freshman}-E_{mountain}\right)}^2}{E_{mountain}}+\frac{{\left(O_{freshman}-E_{city}\right)}^2}{E_{city}}+\frac{{\left(O_{freshman}-E_{\mathrm{hom}e}\right)}^2}{E_{\mathrm{hom}e}}\\ +\frac{{\left(O_{\text{sophomore}}-E_{beach}\right)}^2}{E_{beach}}+\frac{{\left(O_{\text{sophomore}}-E_{mountain}\right)}^2}{E_{mountain}}+\frac{{\left(O_{\text{sophomore}}-E_{city}\right)}^2}{E_{city}}+\frac{{\left(O_{\text{sophomore}}-E_{\mathrm{hom}e}\right)}^2}{E_{\mathrm{hom}e}}\\ +\frac{{\left(O_{\text{junior}}-E_{beach}\right)}^2}{E_{beach}}+\frac{{\left(O_{\text{junior}}-E_{mountain}\right)}^2}{E_{mountain}}+\frac{{\left(O_{\text{junior}}-E_{city}\right)}^2}{E_{city}}+\frac{{\left(O_{\text{junior}}-E_{\mathrm{hom}e}\right)}^2}{E_{\mathrm{hom}e}}\\ +\frac{{\left(O_{\text{senior}}-E_{beach}\right)}^2}{E_{beach}}+\frac{{\left(O_{\text{senior}}-E_{mountain}\right)}^2}{E_{mountain}}+\frac{{\left(O_{\text{senior}}-E_{city}\right)}^2}{E_{city}}+\frac{{\left(O_{\text{senior}}-E_{\mathrm{hom}e}\right)}^2}{E_{\mathrm{hom}e}}\end{array}\end{array}$

The following table represents the chi square test statistic value s given below,

Observed	Expected	$O_i-E_i$	${\left(O_i-E_i\right)}^2$	$\frac{{\left(O_i-E_i\right)}^2}{E_i}$
19	19.7708	-0.7708	0.5942	0.0301
15	15.9688	-0.9688	0.9385	0.0588
18	17.8698	0.1302	0.0170	0.0009
21	19.3906	1.6094	2.5901	0.1336
2	3.5208	-1.5208	2.3129	0.6569
4	2.8438	1.1563	1.3369	0.4701
1	3.1823	-2.1823	4.7624	1.4965
6	3.4531	2.5469	6.4866	1.8785
7	6.2292	0.7708	0.5942	0.0954
3	5.0313	-2.0313	4.1260	0.8201
9	5.6302	3.3698	11.3555	2.0169
4	6.1094	-2.1094	4.4495	0.7283
24	22.4792	1.5208	2.3129	0.1029
20	18.1563	1.8438	3.3994	0.1872
19	20.3177	-1.3177	1.7364	0.0855
20	22.0469	-2.0469	4.1897	0.1900
				${\chi }^2={\displaystyle \sum _{i=1}^n\frac{{\left(O_i-E_i\right)}^2}{E_i}}=8.9517$

So, the value of test statistic, ${\chi }^2$, for a chi- square test for association is 8.9517.

Step 4:

Draw a conclusion and interpret the decision.

Degrees of freedom in a Chi-square test for association:

In a chi-square test for association the number of degrees of freedom for the chi-square distribution of the test statistic is given by,

$df=\left(R-1\right)\cdot \left(C-1\right)$

Where R is the number of rows of the data in the contingency table (not including the row total and C is the number of columns of data in the contingency table (not including the column totals).

Rejection Region for Chi-Square Test for Association:

Reject the null hypothesis, $H_0$, if ${\chi }^2\ge {\chi }_{\alpha }^2$.

From the given information R = 4 and C = 4.

Substitute the above values in the formula of degrees of freedom to get the following,

$\begin{array}{rll}df& =& \left(R-1\right)\cdot \left(C-1\right)\\ & =& \left(4-1\right)\cdot \left(4-1\right)\\ & =& \left(3\right)\cdot \left(3\right)\\ & =& 9\end{array}$

The number of degrees of freedom for this test is 9.

Use the level of significance of $\alpha =0.05$ to find the critical value ${\chi }_{0.05}^2$ of the chi-square distribution with df = 9 in order to make a decision about the null hypothesis.

Use the “Area to the right of the critical value ${\chi }^2$” table to get the following,

${\chi }_{0.05}^2=16.9190$

The test statistic value is ${\chi }^2=8.9517$.

Comparing the test statistic and critical value to get the following,

$8.9517<16.919\iff {\chi }^2<{\chi }_{0.005}^2$. Therefore, we fail to reject the null hypothesis this concludes that age and destination are independent.

So, there is no relationship between age( by college classification) and destination.

Final statement:

There is no relationship between age( by college classification) and destination.