Soc2205 More Midterm Review Questions

Additional Selected Problems and Solutions For Midterm

**Let me know if there are any errors!**

Problem: Healey 1st #2.2 f - j, 2nd #2.2 f - j

• Note: This is a continuation of #2.2 in the first review presentation.

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

Problem: Healey 1st #3.4, 2nd #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 2nd Can. Text)

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

Problems: Healey 1st Can. #3.8, and 2nd Can. #3.8

• For both texts, compute – the mean

– the median

– compare the two

Solutions: Healey 1st Can. #3.8, and 2nd Can. #3.8

• 2000: – Mean = 654.6, Median = 654– Mean>Median. Very slight positive skew.

• 2005: – Mean = 703, Median = 703– Unskewed distribution.

Problem: Healey 1st #3.12, 2nd #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:

22

XN

XS i

Solution: Healey 1st #3.12, 2nd #3.12

• Pretest: Mean = 9.33, Median = 10

• Pretest: s = 3.50

• Posttest: Mean = 12.93, Median = 12

• Posttest: s = 4.40

Problem: Healey 1st #5.2, 2nd #4.2

• Problem Information:

• = 500

• S = 100

• *Remember to draw a “curve” to find the % areas

X

Solution: Healey 1st #5.2, 2nd #4.2

Z score % Area Above % Area Below 1.50 6.68 93.32 -1.00 84.13 15.87 -1.25 89.44 10.56 .86 19.49 80.51- .63 73.57 26.43 .26 39.74 60.26 1.21 11.31 88.69- .02 50.80 49.20 .17 43.25 56.75-1.02 84.61 15.39

Problem: Healey 1st #7.14, 2nd #6.14

• *Compare the widths of the intervals in your answer before looking at the solution on next slide

Solution: Healey 1st #7.14, 2nd #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!

Problem: Healey 1st #7.8, 2nd #6.8

• Problem Information:

• This is the 90% CI– = .10, Z = 1.65

• N = 500

• Calculate Ps first = (50/500) = . 10

Solution: Healey 1st #7.8, 2nd #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%)

Problem: Healey 1st #7.18, 2nd #6.18

• Problem Information:

• This is the 99% CI– = .01, Z = 2.58

• = 3.1 litres/100km

• S = 3.7

• N = 120

X

Solution: Healey 1st #7.18, 2nd #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!