One professor was interested in evaluating the overall performance of students in online sections. He wanted...

One professor was interested in evaluating the overall performance of students in online sections. He wanted to know if students in an online section would do as well as students in a face-to-face section. So one semester he has the same class but one class was delivered in an online format and one section was delivered face-to-face and it served as a control section The grades of each section are listed below ,with 4.0=A down to 0.0=F A. Write out in words the Null and Alternative hypotheses for this situation B.Create a histogram of each class using the bins below. C.What is the correct statistical test to use satistically determine if there is a difference between the delivery methods. Perform the correct Test,highlight the result you use to interpret the test and report in words your conclusion about the hypothesis E. How do these results compare?Practically, what is happening in the classes? On-line: On-line 4.0 4.0 2.3 4.0 4.0 3.7 4.0 0.3 0.0 4.0 4.0 2.0 3.3 4.0 2.0 4.0 3.7 3.3 3.0 2.0 4.0 0.0 3.7 0.0 3.0 0.3 3.0 3.3 3.7 3.3 4.0 3.0 4.0 4.0 4.0 4.0 3.0 0.0 3.3 3.7 0.0 2.3 2.7 3.0 4.0 4.0 4.0 0.0 0.0 3.3 4.0 Lecture: 2.0 3.0 2.3 3.0 3.3 3.0 1.0 4.0 4.0 3.7 3.0 4.0 1.0 2.0 3.0 3.0 1.0 2.0 2.0 1.3 4.0 4.0 4.0 3.7 3.3 1.7 3.7 3.0 4.0 1.7 2.5 3.2 2.9 3.5 3.7 3.5 3.6 3.0 1.5 2.8 Bins: 0.0 0.3 0.7 1.0 1.3 1.7 2.0 2.3 2.7 3.0 3.3 3.7 4.0

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Expert Solution

From both online and lecture data, we observe that both the data do not follow the normal distribution. Hence, the parametric test to compare the online and lecture data is not accepted. Hence, we should use the nonparametric test.

A) Null Hypothesis: The median of online is equal to the median of the lecture

Alternative Hypothesis: The median of online is not equal to the median of the lecture.

C) The correct statistical test to use statistically determine if there is a difference between the delivery methods is the Mann-Whitney Test.

The estimated p-value is 0.2152. Hence, retain the null hypothesis and conclude that the students in an online section do as well as students in a face-to-face section at the 0.05 significance level.

orchestra answered 1 year ago

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