Recent revenue shortfalls in a Midwestern state led to a reduction in the state budget for higher education.To offset the reduction,the largest state university proposed a 25% tuition increase.It was determined that such an increase was needed to simply compensate for the lost support from the state.Random samples of 50 freshmen,50 sophomores,50 juniors,and 50 seniors from the university were asked whether or not they were strongly opposed to the increase,given that it was the minimum increase necessary to maintain the university's budget at current levels.The results are given in the following table. The chi-square statistic for these data equals 21.9.Are the data statistically significant at the 5% significance level under the appropriate null hypothesis?
A)Yes
B)No
C)This cannot be determined from the information given.

Question

Recent revenue shortfalls in a Midwestern state led to a reduction in the state budget for higher education.To offset the reduction,the largest state university proposed a 25% tuition increase.It was determined that such an increase was needed to simply compensate for the lost support from the state.Random samples of 50 freshmen,50 sophomores,50 juniors,and 50 seniors from the university were asked whether or not they were strongly opposed to the increase,given that it was the minimum increase necessary to maintain the university's budget at current levels.The results are given in the following table.  The chi-square statistic for these data equals 21.9.Are the data statistically significant at the 5% significance level under the appropriate null hypothesis?
A)Yes 
B)No 
C)This cannot be determined from the information given.

Quizplus · Accepted Answer

We can use a chi-square test of independence to determine whether there is a significant association between class level and opposition to the tuition increase. The null hypothesis is that there is no association between the two variables.

The expected frequencies for each cell can be calculated by multiplying the row total by the column total and dividing by the overall total. For example, the expected frequency for freshmen who are strongly opposed is (50*200)/200 = 50.

We can use a calculator or a table to find the critical value for chi-square with 3 degrees of freedom and a 5% significance level. The critical value is 7.815.

To calculate the chi-square statistic, we use the formula: 
(observe1 - expect1)^2 / expect1 + (observe2 - expect2)^2 / expect2 + ... + (observe4 - expect4)^2 / expect4

where observe1 is the observed frequency for freshmen who are strongly opposed, expect1 is the expected frequency for freshmen who are strongly opposed, and so on for the other cells.

Using the given data, we get a chi-square statistic of 21.9. This is greater than the critical value of 7.815, so we reject the null hypothesis and conclude that there is a significant association between class level and opposition to the tuition increase.

Recent Revenue Shortfalls in a Midwestern State Led to a Reduction