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# COURSE: JUST 3900 Tegrity Presentation Developed By: Ethan Cooper Final Exam Review.

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COURSE: JUST 3900 Tegrity Presentation Developed By: Ethan Cooper Final Exam Review

Chapter 10: Independent Measures t-test Question 1: One sample from an independent-measures study has n = 5 with SS = 48. The other sample has n = 9 and SS = 32. Question 1: One sample from an independent-measures study has n = 5 with SS = 48. The other sample has n = 9 and SS = 32. a) Compute the pooled variance for the sample. b) Compute the estimated standard error for the mean difference.

Chapter 10: Independent Measures t-test

Question 2: What is the null hypothesis when using an independent measures t-test? Question 2: What is the null hypothesis when using an independent measures t-test?

Chapter 10: Independent Measures t-test

Question 3: An independent measures t-test results in a t-statistic of t = 3.53. Each sample consists of n = 8. Compute the effect size using r 2 ? Question 3: An independent measures t-test results in a t-statistic of t = 3.53. Each sample consists of n = 8. Compute the effect size using r 2 ?

Chapter 10: Independent Measures t-test

Question 4: A researcher is conducting an independent measures t-test (two-tail) with two samples of n = 8. M 1 = 3 and M 2 = 6. The estimated standard error is s (M 1 – M 2 ) = 0.85. Is there a significant treatment effect? Question 4: A researcher is conducting an independent measures t-test (two-tail) with two samples of n = 8. M 1 = 3 and M 2 = 6. The estimated standard error is s (M 1 – M 2 ) = 0.85. Is there a significant treatment effect? (α = 0.05) (α = 0.05)

Chapter 10: Independent Measures t-test

Question 5: The boundaries for a confidence interval are at t = ± 1.753. The sample sizes are n = 7 and n = 10. How confident are we that the unknown mean difference falls in this interval? Question 5: The boundaries for a confidence interval are at t = ± 1.753. The sample sizes are n = 7 and n = 10. How confident are we that the unknown mean difference falls in this interval?

Chapter 10: Independent Measures t-test Question 5 Answer: Question 5 Answer: Find df. Find df. df = 9 + 6 = 15 df = 9 + 6 = 15 Use the t distribution to find the alpha level where t = 1.753 and df = 15 intersect. Use the t distribution to find the alpha level where t = 1.753 and df = 15 intersect. α = 0.10 α = 0.10 An alpha of 0.10 corresponds to 10% in the tails, leaving 90% in the body. An alpha of 0.10 corresponds to 10% in the tails, leaving 90% in the body. 100 – 10 = 90 100 – 10 = 90 Therefore, we are 90% confident that our mean difference falls in this interval. Therefore, we are 90% confident that our mean difference falls in this interval.

Chapter 10: Independent Measures t-test Question 6: A researcher wants to conduct an independent measures t-test, but first, he wants to make sure the homogeneity assumption is not violated. Each sample has n = 8 and the sum of squares for each are SS = 16 and SS = 24. Use an F-Max test to see if the assumption is violated (α = 0.05). Question 6: A researcher wants to conduct an independent measures t-test, but first, he wants to make sure the homogeneity assumption is not violated. Each sample has n = 8 and the sum of squares for each are SS = 16 and SS = 24. Use an F-Max test to see if the assumption is violated (α = 0.05).

Chapter 10: Independent Measures t-test

Chapter 12: ANOVA Question 1: Which of the following F-ratios is most likely to reject the null? Question 1: Which of the following F-ratios is most likely to reject the null? a) F = 1.25 b) F = 1.00 c) F = 2.75 d) F = 4.50 e) None of the Above

Chapter 12: ANOVA Question 1 Answer: Question 1 Answer: D) F = 4.50 D) F = 4.50 F values closer to 1.00 are less likely to fall in the critical region and thus, are less likely to reject the null. F values closer to 1.00 are less likely to fall in the critical region and thus, are less likely to reject the null.

Chapter 12: ANOVA Question 2: Which of the following are possible alternative hypotheses for ANOVA? Question 2: Which of the following are possible alternative hypotheses for ANOVA? a) H 1 : μ 1 ≠ μ 2 = μ 3 b) H 1 : μ 1 ≠ μ 2 ≠ μ 3 c) H 1 : μ 1 = μ 2 ≠ μ 3 d) All of the above e) None of the Above

Chapter 12: ANOVA Question 2 Answer: Question 2 Answer: D) All of the above D) All of the above The alternative hypothesis for ANOVA states that there is a difference between our population means, but it does not identify which means are different. The alternative hypothesis for ANOVA states that there is a difference between our population means, but it does not identify which means are different.

Chapter 12: ANOVA Question 3: What is the advantage of using ANOVA over an independent measures t-test? Question 3: What is the advantage of using ANOVA over an independent measures t-test?

Chapter 12: ANOVA Question 3 Answer: Question 3 Answer: ANOVA can be used when comparing more than two populations without increasing the risk of Type I error. ANOVA can be used when comparing more than two populations without increasing the risk of Type I error.

Chapter 12: ANOVA Question 4: What accounts for between treatments variance? Question 4: What accounts for between treatments variance?

Chapter 12: ANOVA Question 4 Answer: Question 4 Answer: Between treatments variance (MS between ) is caused by both treatment effects and random, unsystematic error. Between treatments variance (MS between ) is caused by both treatment effects and random, unsystematic error.

Chapter 12: ANOVA Question 5: Compute effect size (η 2 ) for a data set with SS between = 70 and SS within = 46. Question 5: Compute effect size (η 2 ) for a data set with SS between = 70 and SS within = 46.

Chapter 12: ANOVA

Question 6: A researcher is using Tukey’s HSD to find which treatments (k = 3 treatments) in his study had an effect. MS within = 3.83 and n = 5. The mean difference between treatments A and B is – 4. Is this a significant mean difference? (α = 0.05) Question 6: A researcher is using Tukey’s HSD to find which treatments (k = 3 treatments) in his study had an effect. MS within = 3.83 and n = 5. The mean difference between treatments A and B is – 4. Is this a significant mean difference? (α = 0.05)

Chapter 12: ANOVA

Question 7: Which of the following is not an assumption required for the independent-measures ANOVA? Question 7: Which of the following is not an assumption required for the independent-measures ANOVA? a) The observations within each sample must be independent. b) The samples must all be the same size. c) The populations from which the samples are selected must be normal. d) The populations from which the samples are selected must have equal variances (homogeneity of variance). e) All of the above are required assumptions.

Chapter 12: ANOVA Question 7 Answer: Question 7 Answer: B) The samples must all be the same size. B) The samples must all be the same size. ANOVA requires the same assumption as independent measures t tests: ANOVA requires the same assumption as independent measures t tests: The observations must be independent. The observations must be independent. The populations variances must all be the same. The populations variances must all be the same. The populations must be normally distributed. The populations must be normally distributed.

Chapter 12: ANOVA Question 8: Question 8: SourcesSSdfMS Between20 Within Total200 n = 16 for each sample k = 3 treatments F =

Chapter 12: ANOVA Question 8 Answer: Question 8 Answer: SourcesSSdfMS Between203 – 1 = 220/2 = 10 Within200 – 20 = 180 47 – 2 = 45180/45 = 4 Total20016*3 – 1 = 47 n = 16 for each sample k = 3 treatments F = 10/4 = 2.50

Chapter 17: Chi-Square Question 1: In what situations would we use nonparametric tests as substitutes for parametric tests? Question 1: In what situations would we use nonparametric tests as substitutes for parametric tests?

Chapter 17: Chi-Square Question 1 Answer: Question 1 Answer: The data do not meet the assumptions needed for a standard parametric test. The data do not meet the assumptions needed for a standard parametric test. The data consist of nominal or ordinal measurements, so that it is impossible to compute standard descriptive statistics such as the mean and standard deviation. The data consist of nominal or ordinal measurements, so that it is impossible to compute standard descriptive statistics such as the mean and standard deviation.

Chapter 17: Chi-Square Question 2: To investigate the phenomenon of “home- team advantage,” a researcher recorded the outcomes from 64 college football games on one weekend in October. Of the 64 games, 42 were won by home teams. Does this result provided enough evidence that home teams win significantly more than would be expected by chance? Assume winning and losing are equally likely events if there is no home-team advantage. Use α = 0.05 Question 2: To investigate the phenomenon of “home- team advantage,” a researcher recorded the outcomes from 64 college football games on one weekend in October. Of the 64 games, 42 were won by home teams. Does this result provided enough evidence that home teams win significantly more than would be expected by chance? Assume winning and losing are equally likely events if there is no home-team advantage. Use α = 0.05

Chapter 17: Chi-Square Question 2 Answer: Question 2 Answer: Step 1: State the hypothesis Step 1: State the hypothesis H 0 : There is no home-team advantage. H 0 : There is no home-team advantage. H 1 : There is a home-team advantage. H 1 : There is a home-team advantage. WinsLosses 4222 f0f0 WinsLosses 50%

Chapter 17: Chi-Square Question 2 Answer: Question 2 Answer: Step 2: Locate the critical region Step 2: Locate the critical region Find df. Find df. df = C – 1 = 2 – 1 = 1 df = C – 1 = 2 – 1 = 1 Use df and alpha level (α = 0.05) to find the critical Χ 2 value for the test. Use df and alpha level (α = 0.05) to find the critical Χ 2 value for the test. Χ 2 crit = 3.84 Χ 2 crit = 3.84

Chapter 17: Chi-Square Question 2 Answer: Question 2 Answer: Step 3: Calculate the Chi-Square Statistic Step 3: Calculate the Chi-Square Statistic Identify proportions required to compute expected frequencies (f e ). Identify proportions required to compute expected frequencies (f e ). The null specifies the proportion for each cell. With a sample of 64 games, the expected frequencies for each category (wins/losses) are equal in proportion (50%). The null specifies the proportion for each cell. With a sample of 64 games, the expected frequencies for each category (wins/losses) are equal in proportion (50%). Calculate the expected frequencies with proportions from H 0. Calculate the expected frequencies with proportions from H 0. f e = pn = (0.50) * (64) = 32 games in each category f e = pn = (0.50) * (64) = 32 games in each category Calculate the Chi-Square Statistic Calculate the Chi-Square Statistic WinsLosses 4222 32 f0f0 fefe

Chapter 17: Chi-Square

Question 2 Answer: Question 2 Answer: Step 4: Make a Decision Step 4: Make a Decision If X 2 ≤ 3.84, fail to reject H 0 If X 2 ≤ 3.84, fail to reject H 0 If X 2 > 3.84, reject H 0 If X 2 > 3.84, reject H 0 6.25 > 3.84, thus, we reject H 0, which means home-team advantage does effect the outcome of college football games. 6.25 > 3.84, thus, we reject H 0, which means home-team advantage does effect the outcome of college football games.

Chapter 17: Chi-Square Question 3: A researcher obtains a sample of 200 high school students. The students are given a description of a psychological research study and asked whether they would volunteer to participate. The researcher also obtains an IQ score for each student and classifies the students into high, medium, and low IQ groups. Do the following data indicate a significant relationship between IQ and volunteering? (α = 0.05) Question 3: A researcher obtains a sample of 200 high school students. The students are given a description of a psychological research study and asked whether they would volunteer to participate. The researcher also obtains an IQ score for each student and classifies the students into high, medium, and low IQ groups. Do the following data indicate a significant relationship between IQ and volunteering? (α = 0.05)

Chapter 17: Chi-Square High IQMedium IQLow IQ Volunteer437334 Not Volunteer72716 150 50 100 50

Chapter 17: Chi-Square Question 3 Answer: Question 3 Answer: Step 1: State the hypothesis Step 1: State the hypothesis H 0 : There is no relationship between volunteering and IQ. H 0 : There is no relationship between volunteering and IQ. H 1 : There is a relationship between volunteering and IQ. H 1 : There is a relationship between volunteering and IQ.

Chapter 17: Chi-Square Question 3 Answer: Question 3 Answer: Step 2: Locate the critical region. Step 2: Locate the critical region. Find df. Find df. df = (R -1)*(C – 1) = (2 – 1)*(3 – 1) = (1)*(2) = 2 df = (R -1)*(C – 1) = (2 – 1)*(3 – 1) = (1)*(2) = 2 Use df (2) and alpha level (0.05) to find the critical X 2 value. Use df (2) and alpha level (0.05) to find the critical X 2 value. X 2 crit = 5.99 X 2 crit = 5.99

Chapter 17: Chi-Square High IQMedium IQLow IQ Volunteer Not Volunteer 150 50 10050

Chapter 17: Chi-Square High IQMedium IQLow IQ Volunteer(37.5)43(75)73(37.5)34 Not Volunteer(12.5)7(25)27(12.5)16

Chapter 17: Chi-Square Question 3 Answer: Question 3 Answer: Step 4: Make a Decision Step 4: Make a Decision If X 2 ≤ 5.99, fail to reject H 0 If X 2 ≤ 5.99, fail to reject H 0 If X 2 > 5.99, reject H 0 If X 2 > 5.99, reject H 0 4.75 > 5.99, thus, we fail to reject H 0, which means there is not a significant relationship between volunteering and IQ. 4.75 > 5.99, thus, we fail to reject H 0, which means there is not a significant relationship between volunteering and IQ.

Chapter 17: Chi-Square Question 4: A researcher completes a chi-square test for independence and obtains X 2 = 6.2 for a sample of n = 40 participants. Question 4: A researcher completes a chi-square test for independence and obtains X 2 = 6.2 for a sample of n = 40 participants. If the frequency data formed a 2 X 2 matrix, what is the phi- coefficient for the test? If the frequency data formed a 2 X 2 matrix, what is the phi- coefficient for the test? If the frequency data formed a 3 X 3 matrix, what is Cramer’s V for the test? If the frequency data formed a 3 X 3 matrix, what is Cramer’s V for the test?

Chapter 17: Chi-Square

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