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November 22, 2014

November 22, 2014

Posted by **Evan** on Monday, November 26, 2012 at 5:34pm.

"A stats instructor is interested in investigating time of day his lectures were held on student' test scores. He selected a sample of 15 students from his morning (n=7) and evening (n=8) statistics classes and recorded their midterm and final exam scores (shown below). He was interested in examining whether students performed better in morning vs. evening classes on their final exam scores. He also suspected poorer performance on the final exam compared to the midterm among those attending the morning class.

Morning Class (n=7):

Midterm | Final

80 | 59

80 | 64

75 | 65

92 | 70

78 | 67

79 | 58

69 | 79

Evening Class (n=8):

Midterm | Final

62 | 40

70 | 45

80 | 55

78 | 55

84 | 60

64 | 70

84 | 60

64 | 70

84 | 60

64 | 70

84 | 60

70 | 63

The instructor was also curious about what type of exam students preferred to receive in the course (multiple choice or short-answer). He observed the morning class, 6 students preferred multiple choice exams. Evening class, 2 students preferred multiple choice. Does he have evidence that 50% of the students preferred multiple choice exams in the evening class? Finally, can the instructor conclude that there is a relation between test type preference and time of class? If yes, please measure the strength of the relation."

- Statistics help? -
**PsyDAG**, Monday, November 26, 2012 at 7:02pmWe do not do your homework for you. Although it might take more effort to do the work on your own, you will profit more from your effort. We will be happy to evaluate your work though.

Are you going to consider the mid and final test scores separately or pool them?

Find the mean first = sum of scores/number of scores

Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.

Standard deviation = square root of variance

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.

I'll let you do the calculations.

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