# Statistics: Test freshmen and juniors for stressful events

Reasearchers were interested in examining the number of stressful events experienced by college students. They collected data from 2 groups of students: freshmen and juniors. The instrument they used to collect the data was a simple checklist that asked the student to check every item that that has occurred to you in the last three months. There were 100 items sample terms such as I moved, I got married, someone close to me died, and I started a new job.

Freshmen data set: 8, 9, 10, 9, 10, 7, 8, 10, 10, 80

1. compute the mean

2. compute the median

4. compute the range

5. compute the standard deviation (you may use a spreadsheet if used include in your work)

6. what is the variance?

7. how would you describe this set of data? in other words how would you describe the average score (mean, median, or mode and why?) How would you describe the variability (range, standard deviation or both? Why?)

Junior data set: 3, 3, 3, 4, 3, 2, 4, 2, 3, 4

8. compute the mean

9. compute the median

10. what is the mode?

11. compute the range

12. compute the standard deviation. if spreadsheet is used must be included

13. what is the variance?

14. how would you describe this data? how would you describe the average score? (mean, median, or mode and why) how would you describe the variability (range, standard deviation or both and why?)

15. putting it all together: do freshmen and juniors report similar numbers of stressful events? How would you say the two groups compare?

https://brainmass.com/statistics/quantative-analysis-of-data/statistics-test-freshmen-juniors-stressful-events-359731

## SOLUTION This solution is **FREE** courtesy of BrainMass!

Your requested analysis and computations are given in the attached file.

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First we compute the mean, median and mode number of stressful events experienced by freshmen.

Mean =

To find the median, we arrange the data in ascending order.

Ascending order = 7 , 8, 8, 9, 9, 10, 10, 10, 10, 80

Median =

The maximum number of scores reported are 10. Therefore, Mode = 10

Variance = = 505.2

Standard Deviation =

Range = 80 - 7 = 73

As the Mean of 16.1 , the median 9.5 and the mode 10 are all apart from each other and are not in coincidence, the given data is not normally distributed.

The score 80 in the given freshmen data appears to be outlier and if this score of 80 is deleted from the given set of freshmen data, there is a possibility that the data will become normally distributed.

Below we compute the mean, median and mode of the freshmen data without taking in account the score 80 of the data set.

Mean =

For median the ascending order of the data set 7 , 8, 8, 9, 9, 10, 10, 10, 10

Median = 9 (the middle score in the ascending order of data set)

Mode = 10 ( maximum number of repeated score in the given data set)

In the above modified data set, the Range = 10 - 7 = 3

Variance =

Standard Deviation =

Since the Mean of 9, Median 9 and Mode of 10 are almost coinciding with each other, the data set without the score of 80 is said to be normally distributed.

Now we find the Mean, Median and Mode of the data set of Juniors.

Mean =

For Median the ascending order of the data = 2, 2, 3, 3, 3, 3, 3, 4, 4, 4

The Median = = 3

Mode = 3 (most repeated score in the data set)

Range = 4 - 2 = 2

Variance =

Standard Deviation =

From the above observations we see that about 95% of the scores lie in the range of 1.33 times the standard deviation from the mean on both sides.

As the mean number of stressful events reported by Freshmen and Juniors are 9 and 3.1 which are very different from each other, we conclude that freshmen and juniors do not report similar numbers of stressful events.

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