1. Soc2205 More Midterm Review Questions Additional Selected Problems and Solutions For Midterm **Let me know if there are any errors!**

2. Problem: Healey 1 st #2.2 f - j, 2 nd #2.2 f - j Note: This is a continuation of #2.2 in the first review presentation.

3. Solutions to #2.2 f - j f. 0.08 g. 25.47% h. 0.68 i. 3.97 business to nursing majors j. 0.06

4. Problem: Healey 1 st #3.4, 2 nd #3.4 Use the data from this question to calculate the appropriate measure of central tendency – i.e. the mean, median or mode - for #3.4 (ignore the instructions on calculating dispersion in the 2 nd Can. Text)

5. Solutions to #3.4 The mean income is $36,400. The modal marital status is "married". Six respondents do not own BMWs, so the modal category is "no". The mean number of years of schooling is 4.6 Note: median was not most appropriate for any of the variables

6. Problems: Healey 1 st Can. #3.8, and 2 nd Can. #3.8 For both texts, compute –the mean –the median –compare the two

7. Solutions: Healey 1 st Can. #3.8, and 2 nd Can. #3.8 2000: –Mean = 654.6, Median = 654 –Mean>Median. Very slight positive skew. 2005: –Mean = 703, Median = 703 –Unskewed distribution.

8. Problem: Healey 1 st #3.12, 2 nd #3.12 Compute the mean, median and standard deviation for the pretest and posttest data in these problems. For the standard deviation use the computational (working) formula:

9. Solution: Healey 1 st #3.12, 2 nd #3.12 Pretest: Mean = 9.33, Median = 10 Pretest: s = 3.50 Posttest: Mean = 12.93, Median = 12 Posttest: s = 4.40

10. Problem: Healey 1 st #5.2, 2 nd #4.2 Problem Information: = 500 S = 100 *Remember to draw a “curve” to find the % areas

11. Solution: Healey 1 st #5.2, 2 nd #4.2 Z score % Area Above % Area Below 1.50 6.6893.32 -1.0084.1315.87 -1.2589.4410.56.8619.4980.51 -.6373.5726.43.2639.7460.26 1.2111.3188.69 -.0250.8049.20.1743.2556.75 -1.0284.6115.39

12. Problem: Healey 1 st #7.14, 2 nd #6.14 *Compare the widths of the intervals in your answer before looking at the solution on next slide

13. Solution: Healey 1 st #7.14, 2 nd #6.14 Sample A (N = 100) : 0.40  0.10 Sample B (N = 1000): 0.40  0.03 Sample C (N = 10,000): 0.40  0.01 *notice how as the sample size N gets larger, the interval width decreases? Larger samples produce a more efficient estimate!

14. Problem: Healey 1 st #7.8, 2 nd #6.8 Problem Information: This is the 90% CI –  =.10,Z =  1.65 N = 500 Calculate Ps first = (50/500) =. 10

15. Solution: Healey 1 st #7.8, 2 nd #6.8 CI = 0.10  0.04 Expand this to: “I am 90% confident that the population proportion of people who were victims of violent crime is between.06 and.14 (6% and 14%)

16. Problem: Healey 1 st #7.18, 2 nd #6.18 Problem Information: This is the 99% CI –  =.01,Z =  2.58 = 3.1 litres/100km S = 3.7 N = 120

17. Solution: Healey 1 st #7.18, 2 nd #6.18 99% CI = 3.1  0.87 The 99% interval is between 2.23 and 3.97 litres/100 km. Manufacturer’s claim is 3.0 litres/100 km. This claim lies within the confidence interval range so the manufacturer is telling the truth!