Chapter 10.1, Problem 10.22E

(a)

To determine

To Explain: the null and alternative hypothesis for the question of interest.

Answer to Problem 10.22E

  $H_0:$ There is no connection between parent smoking habit and Student smoking habit.

  $H_1:$ There is a connection between parent smoking habit and Student smoking habit.

Explanation of Solution

Given:

	Student smokes	Student does not smoke
Both parents smoke	400	1380
One parent smokes	416	1823
Neither parent smokes	188	1168

Suppose

  $\alpha =0.05=5\%$

The null hypothesis statement is that there is no connection between the variables where the alternative hypothesis statement is that there is a connection between the variable.

  $H_0:$ There is no connection between parent smoking habit and Student smoking habit.

  $H_1:$ There is a connection between parent smoking habit and Student smoking habit.

(b)

To determine

To find: the expected cell counts and writes a sentence that explains in simple language about expected counts are.

Answer to Problem 10.22E

  $\begin{array}{c}{E}_{11}=332.4874\\ E_{12}=1447.5126\\ E_{21}=418.2244\\ E_{22}=1820.7756\\ E_{31}=253.2882\\ E_{32}=1102.7118\end{array}$

Explanation of Solution

Given:

	Student smokes	Student does not smoke
Both parents smoke	400	1380
One parent smokes	416	1823
Neither parent smokes	188	1168

Formula used:

  $E=np$

Calculation:

The row and column totals are

	Student smokes	Student does not smoke	Total
Both parents smoke	400	1380	1780
One parent smokes	416	1823	2239
Neither parent smokes	188	1168	1356
Total	1004	4371	5375

The expected frequencies $E$ are

  $\begin{array}{c}{E}_{11}=\frac{r_1\times c_1}{n}=\frac{1780\times 1004}{5375}=332.4874\\ E_{12}=\frac{r_1\times c_2}{n}=\frac{1780\times 4371}{5375}=1447.5126\\ E_{21}=\frac{r_2\times c_1}{n}=\frac{2239\times 1004}{5375}=418.2244\\ E_{22}=\frac{r_2\times c_2}{n}=\frac{2239\times 4371}{5375}=1820.7756\\ E_{31}=\frac{r_3\times c_1}{n}=\frac{1356\times 1004}{5375}=253.2882\\ E_{32}=\frac{r_3\times c_2}{n}=\frac{1356\times 4371}{5375}=1102.7118\end{array}$

Expected counts are the count that based on the row and column totals, when there are no connection the variables.

(c)

To determine

To find: the chi-square statistics, degree of freedom and p-value.

Answer to Problem 10.22E

Degree of freedom is 2

  ${\chi }^2=37.5664$

Explanation of Solution

Given:

	Student smokes	Student does not smoke
Both parents smoke	400	1380
One parent smokes	416	1823
Neither parent smokes	188	1168

Formula used:

  ${\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$

Calculation:

The value of the test- statistic is

  $\begin{array}{c}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\\ =\frac{{\left(400-332.4874\right)}^2}{332.4874}+\frac{{\left(1380-1447.5126\right)}^2}{1447.5126}\\ +\frac{{\left(416-418.2244\right)}^2}{418.2244}+\frac{{\left(1823-1820.7756\right)}^2}{1820.7756}\\ +\frac{{\left(188-253.2882\right)}^2}{253.2882}+\frac{{\left(1168-1102.7118\right)}^2}{1102.7118}\\ =37.5664\end{array}$

The degree of freedom is

  $df=(r-1)\left(c-1\right)=\left(3-1\right)\left(2-1\right)=2$

The P-Value is the probability of getting the value of the test statistic

  $P<0.001$

(d)

To determine

To Explain: the conclusion about significance.

Explanation of Solution

If the P-value is equal or less than to the significance level, then the alternative hypothesis is accepted and null hypothesis is rejected:

  $P<0.05\Rightarrow {\text{Reject H}}_0$

There is not enough evidence to help the claim that there is a connection between parent smoking habit and Student smoking.