15.56 For this exercise, use the UCDAVIS2 dataset. Respondents were categorized as male or female and were asked whether they typically sit in the front, middle, or back of the classroom.

1. Carry out the five steps to determine whether there is a significant relationship between these two variables.

15.57: For this exercise, use the UCDAVIS2 dataset. Two variables that were measured were whether the respondent was left or right-handed and whether the respondent finds it easier to make friends with people of the same or opposite sex. Carry out the five steps to determine whether there is a significant relationship between those variables.

Solution: The following crosstab is obtained:

We are interested in testing the following null and alternative hypotheses:

From the table above we compute the table with the expected values

The way those expected frequencies are calculated is shown below:

\({{E}_{\text{1},\text{1}}}=\frac{{{R}_{\text{1}}}\times {{C}_{\text{1}}}}{T}=\frac{\text{20}\times \text{137}}{\text{234}}=\text{11}\text{.7094}\), \({{E}_{\text{1},\text{2}}}=\frac{{{R}_{\text{1}}}\times {{C}_{\text{2}}}}{T}=\frac{\text{20}\times \text{97}}{\text{234}}=\text{8}\text{.2906}\), \({{E}_{\text{2},\text{1}}}=\frac{{{R}_{\text{2}}}\times {{C}_{\text{1}}}}{T}=\frac{\text{214}\times \text{137}}{\text{234}}=\text{125}\text{.2906}\), \({{E}_{\text{2},\text{2}}}=\frac{{{R}_{\text{2}}}\times {{C}_{\text{2}}}}{T}=\frac{\text{214}\times \text{97}}{\text{234}}=\text{88}\text{.7094}\)

Finally, we use the formula \(\frac{{{\left( O-E \right)}^{2}}}{E}\) to get

The calculations required are shown below:

\(\frac{{{\left( \text{13}-\text{11}\text{.7094} \right)}^{2}}}{\text{11}\text{.7094}}=\text{0}\text{.1422}\), \(\frac{{{\left( \text{7}-\text{8}\text{.2906} \right)}^{2}}}{\text{8}\text{.2906}}=\text{0}\text{.2009}\), \(\frac{{{\left( \text{124}-\text{125}\text{.2906} \right)}^{2}}}{\text{125}\text{.2906}}=\text{0}\text{.0133}\), \(\frac{{{\left( \text{90}-\text{88}\text{.7094} \right)}^{2}}}{\text{88}\text{.7094}}=\text{0}\text{.0188}\)

Hence, the value of Chi-Square statistics is

The critical Chi-Square value for \(\alpha =\text{0}\text{.05}\) and \(\left( \text{2}-1 \right)\times \left( \text{2}-1 \right)=\text{1}\) degrees of freedom is \(\chi _{C}^{2}=\text{3}\text{.841}\). Since \({{\chi }^{2}}=\sum{\frac{{{\left( {{O}_{ij}}-{{E}_{ij}} \right)}^{2}}}{{{E}_{ij}}}}=\text{0}\text{.375}\) < \(\chi _{C}^{2}=\text{3}\text{.841}\), then we fail to reject the null hypothesis, which means that we don't have enough evidence to reject the null hypothesis of independence.

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