Basic Quantitative Methods in the Social Sciences (AKA Intro Stats)

1
Basic Quantitative Methods in the Social
Sciences(AKA Intro Stats)

• 02-250-01
• Lecture 6 - Review

2
Change to Help Clinic Hours

• Help Clinic hours for next week are changed as
  follows
• Tuesday June 10 - 1200 - 500 PM (not 100-400)
• Wednesday, June 11 - 1100 AM - 400 PM (not
  1230-330)
• Thursday, June 12 - NO HELP CLINIC HOURS

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Its Review Time!

• Here are some review questions that will resemble
  exam questions
• Note The answers are here but dont look at
  them until youve completed the questions!

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Problem 1

• Suppose a researcher randomly selects 10 students
  buying sandwiches from the CAW Centre cafeteria
  and asks them to rate the sandwiches on a scale
  from 1 (bad taste) to 10 (great taste) (the
  researcher is wondering how the entire university
  student population rates the sandwiches but cant
  afford to interview the entire student body)
• He obtains the following ratings
• 4, 6, 5, 8, 7, 3, 10, 2, 5, 5
• What are the mean, median, and mode of this data
  set?
• What are the variance and standard deviation of
  this data set?

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Problem 1

• Data set (in order) 2, 3, 4, 5, 5, 5, 6, 7, 8,
  10
• Mean Add all numbers together and divide by n
• (23455567810)/10 55/10 5.5
• Median There are 10 scores, so find scores in
  the 5th and 6th positions, add together, and find
  the average (55)/2 5
• Mode Most frequently occurring score 5

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Problem 1

• Variance and standard deviation
• Decide Are we using a sample or population
  formula? A sample formula!

X X2
2 4
3 9
4 16
5 25
5 25
5 25
6 36
7 49
8 64
10 100
SX55 SX2353
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Problem 2

• Suppose the length of time spent studying for a
  Stats exam is normally distributed with a mean of
  10 hours and a standard deviation of 2 hours.
  (N200)
• A. What proportion of students study for less
  than 7.5 hours?
• B. How many students study for between 11 and 14
  hours?

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Problem 2 A

• A
• From Table E.10, find the area in the smaller
  portion for z -1.25 .1056
• Therefore, a proportion of .1056 students study
  for less than 7.5 hours for their Stats exam

z -1.25 0 X 7.5 10
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Problem 2 B

• B
• From Table E.10, find the area in the mean to z
  for z.50 .1915, and for z2.00 .4772
• Now .4772-.1915.2857
• To find how many (.2857)(200)57.14
• Therefore, approx. 57 students study for between
  11 and 14 hours for their Stats exam

z 0 .50 2.00 X 10 11 14
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Problem 3

• The average University of Windsor student eats
  3000 calories a day with a standard deviation of
  400 calories. Professor X wants to know whether
  students living in Residence eat more than the
  average student. He takes a sample of 36 students
  living in Residence and find that their sample
  mean is x 3175 calories. Test the hypothesis at
  the .05 level.

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Example 3 cont.

• 1. State level of significance - ? 0.05 (what
  is usually used)
• 2. State IV and DV
• IV living location (residence or not)
• DV calories
• 3. Null hypothesis
• Students living in residence eat an equal amount
  of food as does the average U of Windsor student.
• Alternative Hypothesis
• Students living in residence eat more than does
  the average U of Windsor student.

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Example 3 continued

• 4. B/c this hypothesis is directional, we use a
  one-tailed test
• 5. Find the rejection region ? 0.05, so with a
  one-tailed test we want a critical value that
  represents a region of rejection that makes up
  0.05 of the area of the tail. From Table E.10 we
  find that the critical value for z is equal to
  1.64 (and since this is a one-tailed test, we are
  interested in 1.64, and not -1.64).

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Example 3 continued

• This means that zcrit 1.64
• 6. Calculate your statistic

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Example 3 continued

• This means our zobs 2.62
• 7. Compare zcrit to zobs
• Is zobs gt zcrit??
• Yes! 2.62 gt 1.64
• B/c our zobs lies beyond zcrit we say our z-value
  falls into the region of rejection the value of
  zobs is greater than the value of zcrit so we
  choose to reject the Ho
• So we reject the null hypothesis, students in
  residence to in fact eat more than the average
  University of Windsor student.

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Example 4
Lets say the average Canadian earns 40000 each
year with a standard deviation of 4300.
Professor Y wants to know if residents of Windsor
earn less than the average Canadian. She samples
49 Windsor residents and finds that their mean
yearly salary is 38050. Test the hypothesis at
the .01 level.
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Example 4 cont.

• 1. State level of significance - ? 0.01
• 2. State IV and DV
• IV Living location (Windsor or not)
• DV yearly income
• 3. Null hypothesis
• Residents of Windsor earn the same amount per
  year as does the average Canadians
• Alternative Hypothesis
• Residents of Windsor earn less per year than does
  the average Canadian.

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Example 4 continued

• 4. B/c this hypothesis is directional, we use a
  one-tailed test
• 5. Find the rejection region ? 0.01, so with a
  one-tailed test we want a critical value that
  represents a region of rejection that makes up
  0.01 of the area of the tail. From Table E.10 we
  find that the critical value for z is equal to
  2.33 (and since this is a one-tailed test on the
  left tail, we are interested in -2.33 and not
  2.33).

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Example 4 continued

• This means that zcrit -2.33
• 6. Calculate your statistic

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Example 4 continued

• This means our zobs -3.17
• 7. Compare zcrit to zobs
• Is zobs lt zcrit??
• Yes! -3.17 lt -2.33
• B/c our zobs lies beyond zcrit we say our z-value
  falls into the region of rejection the value of
  zobs is less than the value of zcrit so we choose
  to reject the Ho
• So we reject the null hypothesis, residents of
  Windsor earn significantly less per year than
  does the average Canadian.

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For Next Class

• Midterm 1
• Dont forget student ID card, pen, pencil(s),
  eraser, calculator, your textbook!