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 - 12:00 - 5:00 PM (not 1:00-4:00) • Wednesday, June 11 - 11:00 AM - 4:00 PM (not 12:30-3:30) • Thursday, June 12 - NO HELP CLINIC HOURS

3. It’s Review Time! • Here are some review questions that will resemble exam questions • Note: The answers are here – but don’t look at them until you’ve completed the questions!

4. 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 can’t 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?

5. 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: • (2+3+4+5+5+5+6+7+8+10)/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: (5+5)/2 = 5 • Mode: Most frequently occurring score = 5

6. Problem 1 • Variance and standard deviation: • Decide: Are we using a sample or population formula? A sample formula!

7. 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. (N=200) A. What proportion of students study for less than 7.5 hours? B. How many students study for between 11 and 14 hours?

8. 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

9. Problem #2: B • B: • From Table E.10, find the area in the “mean to z” for z=.50 = .1915, and for z=2.00 =.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

10. 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.

11. 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.

12. 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).

13. Example 3 continued • This means that zcrit = +1.64 • 6. Calculate your statistic

14. Example 3 continued • This means our zobs = +2.62 • 7. Compare zcrit to zobs • Is zobs > zcrit?? • Yes! 2.62 > 1.64 • B/c ourzobs 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.

15. Example 4 Let’s 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.

16. 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.

17. 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).

18. Example 4 continued • This means that zcrit = -2.33 • 6. Calculate your statistic

19. Example 4 continued • This means our zobs = -3.17 • 7. Compare zcrit to zobs • Is zobs < zcrit?? • Yes! -3.17 < -2.33 • B/c ourzobs 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.

20. For Next Class • Midterm #1 • Don’t forget: student ID card, pen, pencil(s), eraser, calculator, your textbook!