COURSE: JUST 3900 Tegrity Presentation Developed By: Ethan Cooper

COURSE: JUST 3900Tegrity 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.

a) Compute the pooled variance for the sample.b) Compute the estimated standard error for the mean difference.


Question 1 Answer:

a)


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


Question 2 Answer: = 0


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 r2?


Question 3 Answer:

This is a large effect.


Question 4: A researcher is conducting an independent measures t-test (two-tail) with two samples of n = 8. M1 = 3 and M2 = 6. The estimated standard error is s(M1 – M2) = 0.85. Is there a significant treatment effect?

(α = 0.05)


Question 4 Answer: Critical t = ±2.131 -3.53 < -2.131, therefore we reject the null. There is a

significant effect.


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 Answer: Find df.

df = 9 + 6 = 15 Use the t distribution to find the alpha level where t = 1.753 and

df = 15 intersect. α = 0.10

An alpha of 0.10 corresponds to 10% in the tails, leaving 90% in the body.

100 – 10 = 90 Therefore, we are 90% confident that our mean difference falls

in this interval.


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 Answer: Find the variances.

Find the critical F-Max statistic F-Maxcrit = 4.99

Find F-Max

1.50 < 4.99, therefore the assumption of homogeneity is not violated.

Chapter 12: ANOVA Question 1: Which of the following F-ratios is most likely

to reject the null?a) F = 1.25b) F = 1.00c) F = 2.75d) F = 4.50e) None of the Above

Chapter 12: ANOVA Question 1 Answer:

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.

Chapter 12: ANOVA Question 2: Which of the following are possible

alternative hypotheses for ANOVA?a) H1: μ1 ≠ μ2 = μ3

b) H1: μ1 ≠ μ2 ≠ μ3

c) H1: μ1 = μ2 ≠ μ3

d) All of the abovee) None 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.

Chapter 12: ANOVA Question 3: What is the advantage of using ANOVA over

an independent measures t-test?


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?


Between treatments variance (MSbetween) is caused by both treatment effects and random, unsystematic error.

Chapter 12: ANOVA Question 5: Compute effect size (η2) for a data set with

SSbetween = 70 and SSwithin = 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. MSwithin = 3.83 and n = 5. The mean difference between treatments A and B is – 4. Is this a significant mean difference? (α = 0.05)


Find q. q = 3.77

Find HSD.

Compare HSD to . 4 > 3.30, therefore treatment A is significantly different from

treatment B.

Chapter 12: 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.


B) The samples must all be the same size. ANOVA requires the same assumption as independent

measures t tests: The observations must be independent. The populations variances must all be the same. The populations must be normally distributed.

Chapter 12: ANOVA Question 8:

Sources SS df MSBetween 20WithinTotal 200

n = 16 for each samplek = 3 treatments

F =


Sources SS df MSBetween 20 3 – 1 = 2 20/2 = 10Within 200 – 20 =

18047 – 2 = 45 180/45 = 4

Total 200 16*3 – 1 = 47

n = 16 for each samplek = 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?

Chapter 17: Chi-Square Question 1 Answer:

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.

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


Step 1: State the hypothesis H0: There is no home-team advantage.

H1: There is a home-team advantage.

Wins Losses42 22

f0

Wins Losses50% 50%


Step 2: Locate the critical region Find df.

df = C – 1 = 2 – 1 = 1 Use df and alpha level (α = 0.05) to find the critical Χ2 value for the

test. Χ2

crit = 3.84


Step 3: Calculate the Chi-Square Statistic Identify proportions required to compute expected frequencies (fe).

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 H0. fe = pn = (0.50) * (64) = 32 games in each category

Calculate the Chi-Square StatisticWins Losses

42 22

32 32

f0

fe


Step 3: Calculate the Chi-Square Statistic


Step 4: Make a Decision If X2 ≤ 3.84, fail to reject H0

If X2 > 3.84, reject H0

6.25 > 3.84, thus, we reject H0, 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)

Chapter 17: Chi-SquareHigh IQ Medium IQ Low IQ

Volunteer 43 73 34

Not Volunteer 7 27 16150

50

50 100 50


Step 1: State the hypothesis H0: There is no relationship between volunteering and IQ. H1: There is a relationship between volunteering and IQ.


Step 2: Locate the critical region. Find df.

df = (R -1)*(C – 1) = (2 – 1)*(3 – 1) = (1)*(2) = 2 Use df (2) and alpha level (0.05) to find the critical X2 value.

X2crit = 5.99


Step 3: Calculate the Chi-Square Statistic for Independence Compute expected frequencies where

High IQ Medium IQ Low IQVolunteer

Not Volunteer

150

50

50 100 50


Step 3: Calculate the Chi-Square Statistic for Independence

4.75

High IQ Medium IQ Low IQVolunteer (37.5)43 (75)73 (37.5)34

Not Volunteer (12.5)7 (25)27 (12.5)16


Step 4: Make a Decision If X2 ≤ 5.99, fail to reject H0

If X2 > 5.99, reject H0

4.75 > 5.99, thus, we fail to reject H0, 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 X2 = 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 3 X 3 matrix, what is Cramer’s V

for the test?


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COURSE: JUST 3900 Tegrity Presentation Developed By: Ethan Cooper