Question

A college administrator wanted to know if the proportion of students who request online classes, lab classes, or lecture classes was different at two different campuses. At each campus, the administrator took a random sample of 250 students and asked each student which type of class they preferred. The conditions for the appropriate test were verified, and the chi-square test statistic for the test was calculated to be 4.01 with an associated p-value of 0.1347. If the significance level of the test was alpha = .05, what conclusion should the college administrator make about the proportion of students who request online classes, lab classes, or lecture classes at two different campuses?

A. There is convincing statistical evidence to suggest that the proportion of students who request certain classes is the same at each campus.
B. There is convincing statistical evidence to suggest that the proportion of students who request certain classes is different at each campus.
C. There is not convincing statistical evidence to suggest that the proportion of students who request certain classes is the same at each campus.
D. There is not convincing statistical evidence to suggest that the proportion of students who request certain classes is different at each campus.
E. There is not convincing statistical evidence to prove that the proportion of students who request certain classes is different at each campus.

Answer 1

Answer: B. There is not convincing statistical evidence to suggest that the proportion of students who request certain classes is different at each campus.

Step-by-step explanation: The null hypothesis is that the proportion of students who request certain classes is the same at each campus. Since the p-value is greater than a, there is insufficient evidence to reject the null hypothesis and conclude the alternative hypothesis, that the proportion of students who request certain classes is different at each campus.

Answer 2

There is convincing statistical evidence to suggest that the proportion of students who request certain classes is the same at each campus. ( A )

Explanation:

Given sample size = 250

Test statistic = 4.01

P-value = 0.1347

significance level = 0.05

while using chi-square we are supposed to reject H0, but we cant get the critical value ( Z - value ) because the information isn't available , but we are left with the P-value and this P-value been very low shows that the Alternate hypothesis is not strong enough for us to reject the Null hypothesis.

hence we will accept the Null Hypothesis