Question

a random sample of 120 students were asked if they play a varsity sport. the results were sorted by gender. a chi² test of independence will be performed to determine if there is evidence of an association between gender and sports participation for students at the school. which of the following is the correct conclusion for this test at a 0.05 significance level? there is sufficient evidence of an association between gender and sports participation for students at the school. there is insufficient evidence of an association between gender and sports participation for students at the school. there is sufficient evidence of no association between gender and sports participation for students at the school. there is insufficient evidence of no association between gender and sports participation for students at the school.

Explanation:

Step1: Calculate the expected frequencies

The formula for the expected frequency \(E_{ij}=\frac{(R_i\times C_j)}{N}\), where \(R_i\) is the row total, \(C_j\) is the column total, and \(N\) is the grand - total.
For example, for the cell of male and plays varsity: \(E=\frac{50\times55}{120}\approx22.92\)
For male and does not play varsity: \(E=\frac{50\times65}{120}\approx27.08\)
For female and plays varsity: \(E=\frac{70\times55}{120}\approx32.08\)
For female and does not play varsity: \(E=\frac{70\times65}{120}\approx37.92\)

Step2: Calculate the \(\chi^{2}\) statistic

The formula for the \(\chi^{2}\) statistic is \(\chi^{2}=\sum\frac{(O - E)^{2}}{E}\), where \(O\) is the observed frequency.
For male and plays varsity: \(\frac{(18 - 22.92)^{2}}{22.92}=\frac{(- 4.92)^{2}}{22.92}=\frac{24.2064}{22.92}\approx1.06\)
For male and does not play varsity: \(\frac{(32 - 27.08)^{2}}{27.08}=\frac{(4.92)^{2}}{27.08}=\frac{24.2064}{27.08}\approx0.89\)
For female and plays varsity: \(\frac{(37 - 32.08)^{2}}{32.08}=\frac{(4.92)^{2}}{32.08}=\frac{24.2064}{32.08}\approx0.75\)
For female and does not play varsity: \(\frac{(33 - 37.92)^{2}}{37.92}=\frac{(-4.92)^{2}}{37.92}=\frac{24.2064}{37.92}\approx0.64\)
\(\chi^{2}=1.06 + 0.89+0.75 + 0.64=3.34\)

Step3: Determine the degrees of freedom and critical value

The degrees of freedom \(df=(r - 1)(c - 1)\), where \(r\) is the number of rows and \(c\) is the number of columns. Here \(r = 2\) (plays varsity and does not play varsity) and \(c=2\) (male and female), so \(df=(2 - 1)\times(2 - 1)=1\)
For a significance level of \(\alpha = 0.05\) and \(df = 1\), the critical value from the \(\chi^{2}\) distribution table is \(3.84\)

Step4: Make a decision

Since the calculated \(\chi^{2}=3.34<3.84\) (critical value)

Answer:

There is insufficient evidence of an association between gender and sports participation for students at the school.