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Basic Quantitative Methods in the Social Sciences (AKA Intro Stats) 02-250-01 Lecture 11 (Review – Revised June 18 10:15 PM)

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A Few Things … You need to be able to decide which hypothesis test statistic to use for the exam questions. We have covered (since the last midterm):You need to be able to decide which hypothesis test statistic to use for the exam questions. We have covered (since the last midterm): z-testz-test One-sample t-testOne-sample t-test Two sample t-testsTwo sample t-tests XIndependent Sample t XDependent (Related) Samples t

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A Few Hints z-testz-test XComparing a sample mean with a population mean XSigma is known (so you calculate standard error) One-sample t-testOne-sample t-test XComparing a sample mean with a population mean XSigma is unknown (so you are given or need to calculate sample standard deviation to get estimated standard error)

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A Few Hints … Independent Sample tIndependent Sample t XComparing two sample means XTwo distinct (independent) groups Dependent (Related) Samples tDependent (Related) Samples t XComparing two sample means XScores are pairs (e.g., pre-post test, couples, members of the same family, etc)

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Question #1 Researcher Q knows that regular Canadians spend an average of $50 per month on take- out food. She thinks that students probably spend more money on take out-food per month than regular Canadians. She asks 25 students how much money they spend on take-out food per month, and finds that they spend an average of $75 per month, with a standard deviation of $10.Researcher Q knows that regular Canadians spend an average of $50 per month on take- out food. She thinks that students probably spend more money on take out-food per month than regular Canadians. She asks 25 students how much money they spend on take-out food per month, and finds that they spend an average of $75 per month, with a standard deviation of $10.

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Question #1 A) State the null and alternative hypothesesA) State the null and alternative hypotheses B) What type of test should you use?B) What type of test should you use? C) Test the researcher’s hypothesis at the.01 level of significanceC) Test the researcher’s hypothesis at the.01 level of significance D) Test the researcher’s hypothesis at the.05 level of significance using the Confidence Interval approachD) Test the researcher’s hypothesis at the.05 level of significance using the Confidence Interval approach

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Question 1 A) State the null and alternative hypothesesA) State the null and alternative hypotheses XH o : students spend the same amount of money on take-out food per month as regular Canadians XH a : students spend more money on take-out food per month than regular Canadians B) What type of test should you use?B) What type of test should you use? XWe are comparing a sample mean with a population mean, but do not know sigma: XWe use a ONE-SAMPLE T-TEST

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Question 1 C) Test the researcher’s hypothesis at the.01 level of significanceC) Test the researcher’s hypothesis at the.01 level of significance Alpha =.01 (one-tailed), df = 24, t crit = 2.492Alpha =.01 (one-tailed), df = 24, t crit = 2.492 B/c t obs > t crit, we reject the H oB/c t obs > t crit, we reject the H o Conclusion: students spend more money on take- out food each month than regular CanadiansConclusion: students spend more money on take- out food each month than regular Canadians

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D) Test the researcher’s hypothesis at the.05 level of significance using the Confidence Interval approachD) Test the researcher’s hypothesis at the.05 level of significance using the Confidence Interval approach B/c is outside of the interval, we reject H o :B/c is outside of the interval, we reject H o : Conclusion: students spend more money on take- out food each month than regular CanadiansConclusion: students spend more money on take- out food each month than regular Canadians

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Question #2 Dr. Smith thinks that pet owners are happier than people who do not own pets. She randomly samples 5 dog owners and 6 people who do not own pets and asks them to rate their level of happiness on the “Happy Scale” (from 1 to 50). The pet owners have a mean “Happy Scale” rating of 35 with a standard deviation of 5, while the people who do not own pets have a mean “Happy Scale” rating of 30 with a standard deviation of 3.Dr. Smith thinks that pet owners are happier than people who do not own pets. She randomly samples 5 dog owners and 6 people who do not own pets and asks them to rate their level of happiness on the “Happy Scale” (from 1 to 50). The pet owners have a mean “Happy Scale” rating of 35 with a standard deviation of 5, while the people who do not own pets have a mean “Happy Scale” rating of 30 with a standard deviation of 3.

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Question #2 A) State the null and alternative hypothesesA) State the null and alternative hypotheses B) State the IV, levels of IV, and DVB) State the IV, levels of IV, and DV C) What type of test should you use?C) What type of test should you use? D) Test Dr. Smith’s hypothesis at the.05 level of significanceD) Test Dr. Smith’s hypothesis at the.05 level of significance E) Test Dr. Smith’s hypothesis at the.05 level of significance using the Confidence Interval approachE) Test Dr. Smith’s hypothesis at the.05 level of significance using the Confidence Interval approach

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A) State the null and alternative hypothesesA) State the null and alternative hypotheses XHo: people who own pets are as happy as people who do not own pets XHa: people who own pets are happier than people who do not own pets B) State the IV, levels of IV, and DVB) State the IV, levels of IV, and DV XIV: pet ownership (yes, no) XDV: Happy Scale rating (happiness) C) What type of test should you use?C) What type of test should you use? XIndependent Samples t-test

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D) Test Dr. Smith’s hypothesis at the.05 level of significanceD) Test Dr. Smith’s hypothesis at the.05 level of significance t crit = 1.833 < t obs, reject Ho Conclusion: pet owners are happier than those who do not own pets df = (n 1 + n 2 ) – 2 = 11 - 2 = 9

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E) Test Dr. Smith’s hypothesis at the.05 level of significance using the Confidence Interval approachE) Test Dr. Smith’s hypothesis at the.05 level of significance using the Confidence Interval approach B/c 0 is within the interval, we retain the H o :B/c 0 is within the interval, we retain the H o : Conclusion: pet owners are as happy as people who do not own petsConclusion: pet owners are as happy as people who do not own pets This is a good example of how changing from a one-tailed to a two-tailed test can affect your H o decisionThis is a good example of how changing from a one-tailed to a two-tailed test can affect your H o decision

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Question #3 Dr. Bob has invented a new drug for treating insomnia. He hopes the drug works, but it may actually make people’s insomnia worse (it is so new that he isn’t sure which way the drug will work). Dr. Bob asks 10 people with insomnia to record how many hours of sleep they get without the drug and then has them take the drug and record their number of hours of sleep again. Dr. Bob’s data is on the next slide:Dr. Bob has invented a new drug for treating insomnia. He hopes the drug works, but it may actually make people’s insomnia worse (it is so new that he isn’t sure which way the drug will work). Dr. Bob asks 10 people with insomnia to record how many hours of sleep they get without the drug and then has them take the drug and record their number of hours of sleep again. Dr. Bob’s data is on the next slide:

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Patient Hrs. sleep before drug Hrs. sleep after drug 135 244 325 423 544 653 716 803 935 1026

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Question #3 A) State the null and alternative hypothesesA) State the null and alternative hypotheses B) State the IV, levels of IV, and DVB) State the IV, levels of IV, and DV C) What type of test should you use?C) What type of test should you use? D) Test Dr. Bob’s hypothesis at the.01 level of significanceD) Test Dr. Bob’s hypothesis at the.01 level of significance E) Test Dr. Bob’s hypothesis at the.05 level of significance using the Confidence Interval approachE) Test Dr. Bob’s hypothesis at the.05 level of significance using the Confidence Interval approach

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A) State the null and alternative hypothesesA) State the null and alternative hypotheses XHo: taking the drug will not change the number of hours of sleep that insomniacs get XHa: taking the drug will make insomniacs sleep either more or less than before they took the drug B) State the IV, levels of IV, and DVB) State the IV, levels of IV, and DV XIV: insomnia drug (no drug, drug) XDV: number of hours of sleep per night C) What type of test should you use?C) What type of test should you use? XThese are pairs of scores, so use Related Samples t

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Patient Hrs. sleep before drug Hrs. sleep after drug 135 244 325 423 544 653 716 803 935 1026 D-2 0 -3 0 2 -5 -3 -2 -4 ΣD=-18 D2D2D2D24 0 9 1 0 4 25 9 4 16 ΣD 2 =72

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D) Test Dr. Bob’s hypothesis at the.01 level of significanceD) Test Dr. Bob’s hypothesis at the.01 level of significance Alpha =.01 (two tail), df = 9 (# of pairs –1)Alpha =.01 (two tail), df = 9 (# of pairs –1) Xt crit = +/-3.250 > -2.714, retain H o XConclusion: the drug does not work – it did not change how much sleep the insomniacs get

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E) Test Dr. Bob’s hypothesis at the.05 level of significance using the Confidence Interval approachE) Test Dr. Bob’s hypothesis at the.05 level of significance using the Confidence Interval approach Since 0 is outside the interval, we reject HoSince 0 is outside the interval, we reject Ho Conclusion: the drug affected the number of hours of sleep that insomniacs get (it actually increased their sleep – we can tell this by looking at the sign of D, but you simply need to say that the drug changed hrs of sleep b/c this is a two-tailed test)Conclusion: the drug affected the number of hours of sleep that insomniacs get (it actually increased their sleep – we can tell this by looking at the sign of D, but you simply need to say that the drug changed hrs of sleep b/c this is a two-tailed test)

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Question #4 A telephone service provider has introduced a new long distance calling program. For a flat fee of $25 per month, callers can place as many long distance telephone calls within Canada as they want each month. The phone company thinks that callers will spend more time on the phone each month if they use the new plan. They know that average customers spend 5 hours on the phone each month, with a standard deviation of 45 minutes. A sample of 30 customers using the new plan spend an average of 6 hours on the phone each month.A telephone service provider has introduced a new long distance calling program. For a flat fee of $25 per month, callers can place as many long distance telephone calls within Canada as they want each month. The phone company thinks that callers will spend more time on the phone each month if they use the new plan. They know that average customers spend 5 hours on the phone each month, with a standard deviation of 45 minutes. A sample of 30 customers using the new plan spend an average of 6 hours on the phone each month.

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Question #4 A) State the null and alternative hypothesesA) State the null and alternative hypotheses B) What type of test should you use?B) What type of test should you use? C) Calculate the standard errorC) Calculate the standard error D) Test the phone company’s hypothesis at the.05 level of significanceD) Test the phone company’s hypothesis at the.05 level of significance

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A) State the null and alternative hypothesesA) State the null and alternative hypotheses XHo: Customers using the new long distance phone plan will spend the same amount of time on the phone as regular phone customers XHa: Customers using the new long distance phone plan will spend more time on the phone than regular phone customers B) What type of test should you use?B) What type of test should you use? XComparing a sample with a population, sigma is know, so use a Z-TEST

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C) Calculate the standard errorC) Calculate the standard error D) Test the phone company’s hypothesis at the.05 level of significanceD) Test the phone company’s hypothesis at the.05 level of significance Conclusion: customers using the new long distance plan spend more time on the phone than regular customersConclusion: customers using the new long distance plan spend more time on the phone than regular customers Note!! Sigma is given in minutes, so you need to state it as hours (45 minutes =.75 hours) – the level of measurement of the means and standard deviation must be the same! z crit = 1.64 < 7.30, reject H o

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Question 5 Professor Z hypothesizes that there is relationship between the degree to which people believe in magic and their appreciation for Harry Potter Books. He gives a random sample of 5 people a self-report measure of Belief in Magic that produces scores ranging from 1 (Don’t believe in magic at all) to 10 (Believe very much in magic). He also asks them to rate their appreciation of Harry Potter books on scale from 1 (Harry Potter books are awful) to 10 (Harry Potter books are fabulous). Is there a relationship between belief in magic and appreciation of Harry Potter books?Professor Z hypothesizes that there is relationship between the degree to which people believe in magic and their appreciation for Harry Potter Books. He gives a random sample of 5 people a self-report measure of Belief in Magic that produces scores ranging from 1 (Don’t believe in magic at all) to 10 (Believe very much in magic). He also asks them to rate their appreciation of Harry Potter books on scale from 1 (Harry Potter books are awful) to 10 (Harry Potter books are fabulous). Is there a relationship between belief in magic and appreciation of Harry Potter books?

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Belief in Ratings of Magic scores Harry Potter

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Correlation Arithmetic

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[] [] The Pearson r r =

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[] [] The Pearson r r =

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[] [] The Pearson r r =

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[] [] The Pearson r r =

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[] [] The Pearson r r =

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[] [] The Pearson r r =

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[] [] The Pearson r r =

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[] [] The Pearson r r =

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The Pearson r r =

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The Pearson r r =

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The Pearson r r =

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The Pearson r r = See Table E.2 (p.440) for n - 2 df ( 5 - 2 = 3 df) and an alpha ( ) of.05

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The Pearson r r = The “Critical r” =.878 r =.793 Therefore, we retain the Ho, belief in magic and appreciation of Harry Potter books do not appear to be significantly correlated.

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Question #6 Chi Square Goodness of Fit Test Example (see answer on next page)Chi Square Goodness of Fit Test Example (see answer on next page) 1. Before an Ontario provincial election, political researchers thought that the Liberal Party would receive 45% of the votes, the Reform Party would receive 25%, the NDP would receive 20%, and the PC party would receive 10% of the votes. After the election, they examined 500 random votes and found that the political parties received the following number of votes:1. Before an Ontario provincial election, political researchers thought that the Liberal Party would receive 45% of the votes, the Reform Party would receive 25%, the NDP would receive 20%, and the PC party would receive 10% of the votes. After the election, they examined 500 random votes and found that the political parties received the following number of votes: Liberal Party:200Liberal Party:200 Reform Party:100Reform Party:100 NDP:150NDP:150 PC Party:50PC Party:50

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Question #6 continued 5. A) State the null and alternative hypotheses5. A) State the null and alternative hypotheses 5. B) Test the hypothesis at the.05 level of significance5. B) Test the hypothesis at the.05 level of significance

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Hypotheses Ho: The votes in the election were as predicted by the political researchers (O=E)Ho: The votes in the election were as predicted by the political researchers (O=E) Ha: The votes in the election were distributed differently than as predicted (O does not equal E)Ha: The votes in the election were distributed differently than as predicted (O does not equal E)

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Chi Square Chart LiberalsReformNDPPC O20010015050 E22512510050 O - E-25 500 (O – E) 2 625 25000 (O – E) 2 /E2.77785250 Sum32.7778 Calculate expected frequencies (multiply each % by n = 500)

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Question #6 Continued Chi Square observed = 32.78Chi Square observed = 32.78 DF = k-1 = 4-1 = 3, alpha =.05DF = k-1 = 4-1 = 3, alpha =.05 Chi Square critical = 7.82Chi Square critical = 7.82 B/c Chi Square obs > Chi Square crit, we reject the Ho.B/c Chi Square obs > Chi Square crit, we reject the Ho. Conclusion: The political researchers were wrong. The votes in the election were distributed differently than as predicted.Conclusion: The political researchers were wrong. The votes in the election were distributed differently than as predicted.

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Question #7 Researchers want to know if men and women get sick at different rates when they are exposed to the flu. They contact 100 men and 100 women who have been in contact with someone with the flu over the past week and record how many got sick and how many stayed well. The results are presented in the table below: Men Women Men Women 5545 55 Sick Stayed Well

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Question #7 Continued A) State the hypothesesA) State the hypotheses B) Test the hypothesis at the.01 level of significanceB) Test the hypothesis at the.01 level of significance

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Hypotheses Ho: Gender and getting sick when exposed to the flu are independentHo: Gender and getting sick when exposed to the flu are independent Ha: Gender and getting sick when exposed to the flu are not independent (are dependent)Ha: Gender and getting sick when exposed to the flu are not independent (are dependent)

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Calculating Expected Frequencies and Chi-Square Men Women O: 55 E: 50 O: 45 E: 50 O: 45 E: 50 O: 55 E: 50 Sick Stayed Well.5 +.5 +.5 +.5 = 2.00 100 100 100 200

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Question #7 Continued Chi Square crit, df = 1, alpha =.01 = 6.63, so retain HoChi Square crit, df = 1, alpha =.01 = 6.63, so retain Ho Conclusion: Women and men get sick from the flu at the same rates (gender and getting sick when exposed to the flu are independent)Conclusion: Women and men get sick from the flu at the same rates (gender and getting sick when exposed to the flu are independent)

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Good Luck!!

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