Social Studies 201 – Winter, 2001

Second Midterm Examination – 1:30 to 2:20 p.m., March 19, 2001

Answer any three (3) questions.  Each question has equal value.

1.      Probabilities.

a. For each of the two quotes below, explain which concepts of probability (theoretical, frequency, subjective) appear to be used in the word in bold letters.

b. In the second quote concerning job outlook, identify a pair of events that is either independent or dependent, and explain why they have this characteristic.

In a report on the possibility of faulty hospital tests,

Montreal radiologist Dr. Gaetan Barrette said he experienced the problem first-hand when he told a patient the ultrasound machine he was using in a remote part of Quebec was so unreliable that if the patient had liver cancer, there was only a 50 per cent chance of finding it.  (The Globe and Mail, March 17, 2001, p. A1).

A recent article, “Arts Grads Face Tough Job Outlook” notes:

The adage that it pays to get an education doesn’t apply to arts and culture graduates, according to a new Statistics Canada study, which suggests that for them Canada is a wasteland.

Compared with other graduates, the federal agency says “arts and culture graduates were more likely to be moonlighting, be self-employed, earn lower pay, change employers, and find only temporary work. (National Post, March 17, 2001, p. A10).

2. Binomial and normal approximation to the binomial.

In the Saskatchewan provincial election of September 16, 1999, there were 58 seats of the Legislative Assembly to be decided.  The NDP obtained 29 seats, the Saskatchewan Party obtained 26, the Liberals obtained 3, and other parties received none.   The distribution of popular vote among the parties is shown in Table 1.  Use the binomial distribution, with probability of success being the proportion of the popular vote, to obtain the following.

a. Using the distributions in Table 2, what is the probability that the Saskatchewan Party would obtain at least 26 seats but less than 30 seats?

b. Use the normal approximation to the binomial to determine the probability that the NDP would obtain exactly 29 seats.

c. Cite any reasons why the probability in b. might not represent the chance that the NDP would win exactly 29 seats.

3. Probabilities from a cross-classification table.   Use Table 3.

The data in Table 3 concern the issue of whether there is a connection between homicide offenders and victims.  The author, Wendy C. Regoeczi, examines this issue by looking at the previous conviction record of both offenders and victims and cross-classifying these.  The categories are identified by letters from A to F.

a.  The author notes that the proportion of victims “with previous criminal records is relatively small” (p. 498).  Obtain a probability that demonstrates this.

b. What is the conditional probability of selecting a victim with a violent conviction (E) given that the offender has no record (A).

c. Is the event of a victim having a violent conviction independent of the event of the offender having no record?

d. Is the event of a victim having a violent conviction close to independent of the event of an offender having a violent conviction?

e. From c. and d. comment on the author’s statement that “these findings provide a relatively weak support for the hypothesis that there is a link between victimization and offending” (p. 499).

4. Normal distribution.

a. As noted in Table 5, the mean minutes of media use daily for Saskatchewan adults is 207 and the standard deviation is 162 minutes daily.  Assuming that use of the media is a normally distributed variable, obtain the following:

i. The proportion of adults who use the media more than 240 minutes daily.

ii. The probability that a randomly selected Saskatchewan adults uses the media between 120 and 180 minutes daily.

iii. If a sample of 631 Saskatchewan adults is selected randomly, how many would be expected to use the media less than 1 hour daily?

b. Compare the results of a. with the numbers or percentages in Table 4.  Explain whether or not daily media use appears to be a normally distributed variable.

5. Interval Estimates.  Use Table 5.

a. Obtain 90% interval estimates for the true mean minutes of media use daily by those aged 15-24 and by those aged 55-64.  From these, does it seem likely that 55-64 year olds use the media more than do 15-24 year olds?

b. Using the data in the last row of Table 5, obtain a 99% interval estimate for the true mean media use daily for all Saskatchewan adults.  Explain why the width of this interval differs from the width of the intervals in a.

Table 3.  Number of offenders and victims of homicide among those aged 12-17, Canada, 1991 to 1995.

 Victim’s Criminal Record Total None (D) Violent Conviction (E) Other Conviction (F) Offender’s Criminal Record None (A) 51 4 7 62 Violent Conviction (B) 25 5 12 42 Other Conviction (C) 19 5 17 41 Total 95 14 36 145

Source: Wendy C. Regoeczi, “Adolescent violent victimization and offending: Assessing the extent of the link,” Canadian Journal of Criminology, October, 2000, p. 500.

Table 4.  Distribution of time spent daily by Saskatchewan respondents using media.

 Minutes per day Number of Respondents Per Cent of Respondents 0-60 99 15.7 60-120 121 19.1 120-180 110 17.4 180-240 77 12.2 240-300 67 10.6 300-360 59 9.4 360-420 32 5.1 420 plus 66 10.5 Total 631 100.0

Table 5.   Sample sizes, means, and standard deviations of minutes spent daily using media by Saskatchewan respondents of various ages.

 Age Sample Size Mean Minutes Daily Standard Deviation of Minutes Daily 15-24 80 168 128 25-34 113 154 123 35-44 115 180 137 45-54 84 179 133 55-64 88 202 160 65-74 81 280 183 75-84 51 336 195 85+ 19 334 239 Total 631 207 162

Source for Tables 4 and 5.  Statistics Canada, General Social Survey, Cycle 12: Time Use, 1998.

Table 1.  Popular Vote by Party, Saskatchewan Election, 1999.

 Party Per Cent of Popular Vote NDP 38.7% Saskatchewan  Party 39.6% Liberal Party 20.2% New Green Alliance 1.5%

Table 2.  Binomial probability distributions.

 n=58 p=0.387 n=58 p=0.396 n=58 p=0.202 n=58 p=0.0152 K    P( X = K) 8    0.0000 9    0.0001 10    0.0002 11    0.0007 12    0.0017 13    0.0038 14    0.0076 15    0.0141 16    0.0240 17    0.0374 18    0.0538 19    0.0714 20    0.0880 21    0.1005 22    0.1067 23    0.1054 24    0.0971 25    0.0833 26    0.0668 27    0.0500 28    0.0349 29    0.0228 30    0.0139 31    0.0079 32    0.0042 33    0.0021 34    0.0010 35    0.0004 36    0.0002 37    0.0001 38    0.0000 K    P( X = K) 9    0.0000 10    0.0002 11    0.0004 12    0.0011 13    0.0026 14    0.0055 15    0.0106 16    0.0186 17    0.0301 18    0.0450 19    0.0621 20    0.0794 21    0.0942 22    0.1039 23    0.1066 24    0.1019 25    0.0909 26    0.0756 27    0.0588 28    0.0427 29    0.0289 30    0.0183 31    0.0109 32    0.0060 33    0.0031 34    0.0015 35    0.0007 36    0.0003 37    0.0001 38    0.0000 K   P( X = K) 1    0.0000 2    0.0002 3    0.0010 4    0.0036 5    0.0099 6    0.0221 7    0.0415 8    0.0669 9    0.0941 10    0.1167 11    0.1289 12    0.1278 13    0.1145 14    0.0932 15    0.0692 16    0.0471 17    0.0294 18    0.0170 19    0.0090 20    0.0045 21    0.0020 22    0.0009 23    0.0003 24    0.0001 25    0.0000 K   P( X = K) 0    0.4113 1    0.3682 2    0.1620 3    0.0467 4    0.0099 5    0.0017 6    0.0002 7    0.0000