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: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 questionsHere 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!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)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: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 mean, median, and mode of this data set? What are the variance and standard deviation 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, 10Data set (in order): 2, 3, 4, 5, 5, 5, 6, 7, 8, 10 Mean: Add all numbers together and divide by n: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 5 th and 6 th positions, add together, and find the average: (5+5)/2 = 5Median: There are 10 scores, so find scores in the 5 th and 6 th positions, add together, and find the average: (5+5)/2 = 5 Mode: Most frequently occurring score = 5Mode: Most frequently occurring score = 5

6. Problem 1 Variance and standard deviation:Variance and standard deviation: Decide: Are we using a sample or population formula? A sample formula!Decide: Are we using a sample or population formula? A sample formula! X X2X2X2X2 24 39 416 525 525 525 636 749 864 10100 ΣX=55 ΣX 2 =353

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)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: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-1.250 X 7.5 10

9. Problem #2: B 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 z0.50 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.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)1. State level of significance -  = 0.05 (what is usually used) 2. State IV and DV2. State IV and DV  IV = living location (residence or not)  DV = calories 3. Null hypothesis:3. Null hypothesis:  Students living in residence eat an equal amount of food as does the average U of Windsor student. Alternative Hypothesis: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 test4. 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).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 z crit = +1.64This means that z crit = +1.64 6. Calculate your statistic6. Calculate your statistic

14. Example 3 continued This means our z obs = +2.62This means our z obs = +2.62 7. Compare z crit to z obs7. Compare z crit to z obs Is z obs > z crit ??Is z obs > z crit ??  Yes! 2.62 > 1.64  B/c our z obs lies beyond z crit we say our z-value falls into the region of rejection: the value of z obs is greater than the value of z crit so we choose to reject the H o  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 -  = 0.011. State level of significance -  = 0.01 2. State IV and DV2. State IV and DV  IV = Living location (Windsor or not)  DV = yearly income 3. Null hypothesis: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 test4. 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).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 z crit = -2.33This means that z crit = -2.33 6. Calculate your statistic6. Calculate your statistic

19. Example 4 continued This means our z obs = -3.17This means our z obs = -3.17 7. Compare z crit to z obs7. Compare z crit to z obs Is z obs < z crit ??Is z obs < z crit ??  Yes! -3.17 < -2.33  B/c our z obs lies beyond z crit we say our z-value falls into the region of rejection: the value of z obs is less than the value of z crit so we choose to reject the H o  So: we reject the null hypothesis, residents of Windsor earn significantly less per year than does the average Canadian.

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