Children at home?Observed College Major Yes No Totals Applied Behavioral Science 301040 Criminal Justice 10717 Political Science 101020 Psychology 321143 Totals 8238129χ2=5.37 1. How many variables are involved in this chi-square test?2. Which type of chi-square test is this? Goodness of Fit or Test of Independence3. How many degrees of freedom are involved?4. Using the chi-square critical values table, is the result of this test statistically significant? Yes or No

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Children at home?Observed College Major  Yes  No  Totals  Applied Behavioral Science 301040 Criminal Justice 10717 Political Science 101020 Psychology 321143 Totals 8238129χ2=5.37 1. How many variables are involved in this chi-square test?2. Which type of chi-square test is this? Goodness of Fit or Test of Independence3. How many degrees of freedom are involved?4. Using the chi-square critical values table, is the result of this test statistically significant? Yes or No

Balraj Conrad · Accepted Answer

Given Chi-square contingency table:1) There are two variables involved in this Ch-square test and they are: &#039;College major&#039; and &#039;Children at Home&#039;2) Generally there are two types of Chi-square tests:A chi-square goodness of fit test is used to determine if sample data matches a population.A chi-square test for independence compares two variables in a contingency table to see if they are dependent on each other i.e., to check if they are related.So, this is a A chi-square test for independence where, we have a contingency table with two variables.And this test is used to see if there is any relationship between the two variables &quot;College major&quot; and &quot;children at home&quot;3) Degrees of freedom: It is the number of categories minus one.For a contingency table having &#039;r&#039; rows and &#039;c&#039; columns, degrees of freedom are obtained as follows:r : number of rowsc : number of columns.D=F(r−1)(c−1)=(4−1)(2−1)=3Therefore, degrees of freedom involved in this test is 3.

mapogoosae · Accepted Answer

4) State the hypothesis as follows:Null Hypothesis: &#039;College Major&#039; and &#039;Children at home&#039; are independent.Alternative Hypothesis: &#039;College Major&#039; and &#039;Children at home&#039; are not independent (dependent)Using a chi-square table, look up for 3 degrees of freedom in the row and 0.05 (usually significance level α is 0.05) significance level along the column, the intersection of these values is 7.815Given computed value of chi-square statistic is 5.37.Since, the critical value at 5% significance level is higher than the computed value, we cannot reject the null hypothesis.p-value using calculator is 0.146624 and is greater than significance level 0.05. The result is not significant at p &amp;lt; 0.05Therefore, the result of the test is statistically insignificant.

\(\begin{array}{c} &\text{Children at home?} \\ \text {Observed} &\begin{array}{|c|c|c|} \hline

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