Question

a study is conducted in a large high school to determine if the proportion of students who plan to come to the upcoming football game differs across the grades. a random sample of 50 ninth graders, 50 tenth graders, 50 eleventh graders, and 50 twelfth graders was selected. the responses are displayed in the table. the principal would like to know if these data provide convincing evidence that the distribution of responses differs across the grades in the population of all students in her school. the random and 10% conditions are met, and this table of expected counts shows that all expected counts are at least 5. what is the value of the chi - square test statistic for this test?

Explanation:

Step1: Recall the chi - square formula

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.

Step2: Calculate \((O - E)^{2}/E\) for each cell

• For 9th grade, yes: \(\frac{(24 - 26.25)^{2}}{26.25}=\frac{(- 2.25)^{2}}{26.25}=\frac{5.0625}{26.25}=0.19\)
• For 9th grade, no: \(\frac{(26 - 23.75)^{2}}{23.75}=\frac{(2.25)^{2}}{23.75}=\frac{5.0625}{23.75}\approx0.213\)
• For 10th grade, yes: \(\frac{(28 - 26.25)^{2}}{26.25}=\frac{(1.75)^{2}}{26.25}=\frac{3.0625}{26.25}\approx0.117\)
• For 10th grade, no: \(\frac{(22 - 23.75)^{2}}{23.75}=\frac{(-1.75)^{2}}{23.75}=\frac{3.0625}{23.75}\approx0.129\)
• For 11th grade, yes: \(\frac{(26 - 26.25)^{2}}{26.25}=\frac{(-0.25)^{2}}{26.25}=\frac{0.0625}{26.25}\approx0.0024\)
• For 11th grade, no: \(\frac{(24 - 23.75)^{2}}{23.75}=\frac{(0.25)^{2}}{23.75}=\frac{0.0625}{23.75}\approx0.0026\)
• For 12th grade, yes: \(\frac{(27 - 26.25)^{2}}{26.25}=\frac{(0.75)^{2}}{26.25}=\frac{0.5625}{26.25}=0.022\)
• For 12th grade, no: \(\frac{(23 - 23.75)^{2}}{23.75}=\frac{(-0.75)^{2}}{23.75}=\frac{0.5625}{23.75}\approx0.024\)

Step3: Sum up all the values

\(\chi^{2}=0.19 + 0.213+0.117 + 0.129+0.0024 + 0.0026+0.022+0.024\)
\(\chi^{2}=0.7\)

Answer:

\(0.7\)