How are the smoking habits of students related to their parents' smoking? Here is a two-way table from a survey of student s in eight Arizona high schools: begin{array}{c|c}&text{Student smokes}&text{Student does not smoke}&text{Total}hlinetext{Both parents smoke}&400&1380&400+1380=1780hlinetext{One parent smokes}&416&1823&416+1823=2239hlinetext{Neither parent smokes}&188&1168&188+1168=1356hlinetext{Total}&400+416+188=1004&1380+1823+1168=4371&1004+4371=5375end{array} (a) Write the null and alternative hypotheses for the question of interest. (b) Find the expected cell counts. Write a sentence that explains in simple language what "expected counts" are. (c) Find the chi-square statistic, its degrees of freedom, and the P-value. (d) What is your conclusion about significance?

How are the smoking habits of students related to their parents' smoking? Here is a two-way table from a survey of student s in eight Arizona high schools: begin{array}{c|c}&text{Student smokes}&text{Student does not smoke}&text{Total}hlinetext{Both parents smoke}&400&1380&400+1380=1780hlinetext{One parent smokes}&416&1823&416+1823=2239hlinetext{Neither parent smokes}&188&1168&188+1168=1356hlinetext{Total}&400+416+188=1004&1380+1823+1168=4371&1004+4371=5375end{array} (a) Write the null and alternative hypotheses for the question of interest. (b) Find the expected cell counts. Write a sentence that explains in simple language what "expected counts" are. (c) Find the chi-square statistic, its degrees of freedom, and the P-value. (d) What is your conclusion about significance?

Question
Two-way tables
asked 2021-03-04
How are the smoking habits of students related to their parents' smoking? Here is a two-way table from a survey of student s in eight Arizona high schools:
\(\begin{array}{c|c}&\text{Student smokes}&\text{Student does not smoke}&\text{Total}\\\hline\text{Both parents smoke}&400&1380&400+1380=1780\\\hline\text{One parent smokes}&416&1823&416+1823=2239\\\hline\text{Neither parent smokes}&188&1168&188+1168=1356\\\hline\text{Total}&400+416+188=1004&1380+1823+1168=4371&1004+4371=5375\end{array}\)
(a) Write the null and alternative hypotheses for the question of interest.
(b) Find the expected cell counts. Write a sentence that explains in simple language what "expected counts" are.
(c) Find the chi-square statistic, its degrees of freedom, and the P-value.
(d) What is your conclusion about significance?

Answers (1)

2021-03-05

Let us assume:
\(\alpha=0.05=5\%\)
(a) The null hypothesis states that there is no association between the variables, while the alternative hypothesis states that there is an association between the variables.
\(H_0:\) There is no association between student smoking habit and parent smoking habit
\(H_{\alpha}:\) There is no association between student smoking habit and parent smoking habit
(b) Determine the row and column totals of the given table:
\(\begin{array}{c|c}&\text{Student smokes}&\text{Student does not smoke}&\text{Total}\\\hline\text{Both parents smoke}&400&1380&400+1380=1780\\\hline\text{One parent smokes}&416&1823&416+1823=2239\\\hline\text{Neither parent smokes}&188&1168&188+1168=1356\\\hline\text{Total}&400+416+188=1004&1380+1823+1168=4371&1004+4371=5375\end{array}\)
The expected frequencies E are the product of the column and row total, divided by the table total.
\(E_{11}=\frac{r_1\times c_1}{n}=\frac{1780\times 1004}{5375}\approx332.4874\)
\(E_{12}=\frac{r_1\times c_2}{n}=\frac{1780\times4371}{5375}\approx1447.5126\)
\(E_{21}=\frac{r_2\times c_1}{n}=\frac{2239\times1004}{5375}\approx418.2244\)
\(E_{22}=\frac{r_2\times c_2}{n}=\frac{2239\times4371}{5375}\approx1820.7756\)
\(E_{31}=\frac{r_3\times c_1}{n}=\frac{1356\times1004}{5375}\approx253.2882\)
\(E_{32}=\frac{r_3\times c_2}{n}=\frac{1356\times4371}{5375}\approx1102.7118\)
Expected counts are the counts that we expect based on the row and column totals, when there is no association between the variables.
(c) The chi-square subtotals are the squared differences between the observed abd expected frequencies, divivded by the expected frequency.
The value of the test-statistic is then the sum of the chi-square subtotals:
\(X^2=\sum\frac{(O-E)^2}{E}\)
\(=\frac{(400-322.4874)^2}{332.4874}+\frac{(1380-1447.5126)^2}{1447.5126}+\frac{(416-418.2244)^2}{418.2244}+\frac{(1823-1820.7756)^2}{1820.7756}+\frac{(188-253.2882)^2}{253.2882}+\frac{1168-1102.7119)^2}{1102.7118}\)
The degrees of freedom is the product od the number of row and the number of columns, both decreased by 1.
\(df=(r-1)(c-1)=(3-1)(2-1)=2\)
The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of the chi-square distribution table in the appendix containing the \(X^2\) -value in the row \(df=2:\)
\(P<0.001\)
(d) If the P-value is less than or equal to the significance level, then the null hypothesis is rejected:
\(P<0.05\Rightarrow\text{Reject }H_0\)
There is sufficient evidence to support the claim that there is an association between student smoking habit and parent smoking habit.
Result: (a)
\(H_0:\) There is no association between smoking habit and parent smoking habit.
\(H_{\alpha}:\) There is an association between student smoking habit and parent smoking habit.
(b) 332.4874, 1447.5126, 418.2244, 1820.7756, 253.2882, 1102.7118
Expected counts are the counts that we expect based on the row and column totals, when there is no association between the variables.
(c) \(X^2=37.5664\), degrees of freedom, P<0.01
(d) There is sufficient evidence to support the claim that there is an association between student smoking habit and parent smoking habit.

0

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