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Transcript of Soc2205 More Midterm Review Questions Additional Selected Problems and Solutions For Midterm **Let...
![Page 1: Soc2205 More Midterm Review Questions Additional Selected Problems and Solutions For Midterm **Let me know if there are any errors!**](https://reader036.fdocuments.in/reader036/viewer/2022082713/5697c0101a28abf838ccb354/html5/thumbnails/1.jpg)
Soc2205 More Midterm Review Questions
Additional Selected Problems and Solutions For Midterm
**Let me know if there are any errors!**
![Page 2: Soc2205 More Midterm Review Questions Additional Selected Problems and Solutions For Midterm **Let me know if there are any errors!**](https://reader036.fdocuments.in/reader036/viewer/2022082713/5697c0101a28abf838ccb354/html5/thumbnails/2.jpg)
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.
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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
![Page 4: Soc2205 More Midterm Review Questions Additional Selected Problems and Solutions For Midterm **Let me know if there are any errors!**](https://reader036.fdocuments.in/reader036/viewer/2022082713/5697c0101a28abf838ccb354/html5/thumbnails/4.jpg)
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)
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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
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Problems: Healey 1st Can. #3.8, and 2nd Can. #3.8
• For both texts, compute – the mean
– the median
– compare the two
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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.
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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
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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
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Problem: Healey 1st #5.2, 2nd #4.2
• Problem Information:
• = 500
• S = 100
• *Remember to draw a “curve” to find the % areas
X
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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
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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
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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!
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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
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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%)
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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
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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!