Question

Perform the five step hypothesis testing.

Y? IV. (5 points) A course can be taken for credit either by attending lecture sessions at fixed times and days, or by doing online sessions that can be done at the student's own pace and at those times the student chooses. The course coordinator wants to determine if these two ways of taking the course resulted in a significant difference in achievement as measured by the final exam for the course. The below table gives the scores on an examination with 45 possible points for one group of 9 students who took the course online, and a second group of 9 students who took the course with conventional lectures. At 5% level of significance, do these data present sufficient evidence to indicate that the average grade for students who take the course online is significantly higher than for those who attend a conventional class? Note that preliminary data analyses indicate that we may reasonably assume this data is normally distributed and some computational results are given in the table. Perform five steps of an appropriate hypotheses testing we learned during the lecture Online (X) Classroom (Y) X 1024 1225 1369 961 1225 784 1681 1156 1936 1600 1225 729 961 1024 1156 961 Sum 317 284 11361 9122 841 625 31

Answer 1

Null hypothesis: Ho: 4 = 42 That is, the average grade for students who take the course online is same as for those who attend a conventional class. Alternative hypothesis: H4> 2 That is, the average grade for students who take the course online is higher than for those who attend a conventional class.

Rejection rule: If p-value < a(=0.05), then reject the null hypothesis H. EXCEL (Data Analysis) output: t-Test: Two-Sample Assuming Unequal Variances Online Classroom 35.22222 31.55556 24.44444 20.02778 Mean Variance Observations Hypothesized Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail 16 1.649485 0.05927 1.745884 0.11854 2.119905 The value of the test statistic is t=1.65. Critical value = 1.746. p-value = 0.0593.

Conclusion: Use a significance level, a = 0.05 Here, p-value is 0.0593, which is greater than the value of a = 0.05. That is, p-value (=0.0593) > a{=0.05). Therefore, by the rejection rule, do not reject the null hypothesis H. Interpretation: There is insufficient evidence at level of significance a = 0.05. We cannot conclude that the average grade for students who take the course online is higher than for those who attend a conventional class.