Question

a guidance counselor selected random samples of 25 ninth graders, 25 tenth graders, 25 eleventh graders, and 25 twelfth graders from her large high school. she asked each student whether they have a job. the results are displayed in the table.

she decides to test these hypotheses:
( h_{0} ): there is no difference in the distribution of job status for all 9th, 10th, 11th, and 12th graders at this high school.
( h_{a} ): there is a difference in the distribution of job status for all 9th, 10th, 11th, and 12th graders at this high school.

the conditions for inference are met. the chi - square test statistic is ( chi^{2}=22.21 ). what conclusion should be made? use ( alpha = 0.05 ).

Explanation:

Step1: Determine the degrees of freedom

For a chi - square test of homogeneity (comparing distributions across groups), the degrees of freedom formula is \(df=(r - 1)(c - 1)\), where \(r\) is the number of rows (job status: yes/no, so \(r = 2\)) and \(c\) is the number of columns (grade levels: 9th, 10th, 11th, 12th, so \(c=4\)).

$$df=(2 - 1)\times(4 - 1)=3$$

Step2: Find the critical value

Using a significance level \(\alpha = 0.05\) and \(df = 3\), we look up the critical value in the chi - square distribution table. The critical value \(\chi_{0.05,3}^2=7.815\)

Step3: Compare the test statistic and the critical value

We are given \(\chi^{2}=22.21\). Since \(22.21>7.815\) (i.e., the test statistic is greater than the critical value)

Answer:

We reject the null hypothesis \(H_{0}\). There is sufficient evidence at the \(\alpha = 0.05\) significance level to conclude that there is a difference in the distribution of job status for 9th, 10th, 11th, and 12th graders at this high school.