Question

a survey is conducted to determine if there is a difference in the proportion of students, parents, and teachers who volunteer at least once a month. to investigate, a random sample of 45 students, 25 parents, and 12 teachers was selected from a large high school. the data are displayed in the tables. observed counts: volunteer? yes no student 24 21 parent 12 13 teacher 6 6 expected counts: volunteer? yes no student 23.05 21.95 parent 12.80 12.20 teacher 6.15 5.85 the researcher would like to test these hypotheses: the conditions for inference are met. what is the value of the chi - square test statistic and what are the degrees of freedom for this test? ( chi^{2}=0.19 ), df = 2 ( chi^{2}=0.19 ), df = 5 ( chi^{2}=0.44 ), df = 2 ( chi^{2}=0.44 ), df = 5

Explanation:

Step1: Calculate chi - square test statistic

The formula for the chi - square test statistic is \(\chi^{2}=\sum\frac{(O - E)^{2}}{E}\), where \(O\) is the observed count and \(E\) is the expected count.
For students (yes): \(\frac{(24 - 23.05)^{2}}{23.05}=\frac{0.9025}{23.05}\approx0.0391\)
For parents (yes): \(\frac{(12 - 12.80)^{2}}{12.80}=\frac{0.64}{12.80}=0.05\)
For teachers (yes): \(\frac{(6 - 6.15)^{2}}{6.15}=\frac{0.0225}{6.15}\approx0.0037\)
For students (no): \(\frac{(21 - 21.95)^{2}}{21.95}=\frac{0.9025}{21.95}\approx0.0411\)
For parents (no): \(\frac{(13 - 12.20)^{2}}{12.20}=\frac{0.64}{12.20}\approx0.0525\)
For teachers (no): \(\frac{(6 - 5.85)^{2}}{5.85}=\frac{0.0225}{5.85}\approx0.0038\)
Sum these values: \(0.0391+0.05 + 0.0037+0.0411+0.0525+0.0038 = 0.19\)

Step2: Calculate degrees of freedom

The formula for degrees of freedom in a test of homogeneity (or independence in a contingency table) is \(df=(r - 1)(c - 1)\), where \(r\) is the number of rows and \(c\) is the number of columns. Here \(r = 2\) (yes/no) and \(c=3\) (student/parent/teacher). So \(df=(2 - 1)\times(3 - 1)=2\)

Answer:

\(\chi^{2}=0.19\), \(df = 2\)