I need help with a statistics test

1. In a study that examines whether client self-reports of satisfaction with counseling differ between male and female therapists, when should a researcher use a two-tailed test of significance?

a. The researcher finds that the two variables (gender and client satisfaction with counseling), which generally are thought of being related, are really not related.

b. The researcher believes that she knows the direction of the relationship between the two variables.

c. The researcher isn’t sure whether male or female therapists perform better in therapy.

d. If the client sample is less than 25, the research should use a two-tailed test of

significance.

2. To use the two sample t-test for independent groups:

a. both variables must be at least interval level.

b. one variable must be at least interval level; the other, nominal level.

c. both variables should be considered nominal level.

d. one variable must be considered ordinal level; the other should be interval level.

3. A researcher wishes to compare the "attitudes toward statistics" (regarded as interval level) of a class of 25 students before taking a statistics course with their attitudes after completing the course. An appropriate statistical test would be:

a. one sample t-test

b. paired comparison (dependent samples) t-test

c. the two sample t-test for independent groups.

d. one way ANOVA.

4. The t-test for independent samples is well-suited (appropriate) to compare two treatment intervention methods because:

a. one variable need only be nominal level.

b. the total sample size can be small.

c. subsamples need not be equal in size.

d. all of the above.

5. To compare the highest education level completed for three or more religious groups, a good choice for statistical analysis might be:

a. two way ANOVA.

b. one factor MANOVA.

c. one way ANOVA.

d. t-test for independent groups.

6. To compare the highest education level completed for three or more religious groups, a good choice for statistical analysis might be:

a. two way ANOVA.

b. one factor MANOVA.

c. one way ANOVA.

d. t-test for independent groups.

7. In using t-tests, "degrees of freedom" are a function of:

a. the number of cases in the sample or samples.

b. the level of measurement of the dependent variable.

c. the number of value categories used for the independent variable.

d. all of the above.

Use the below t-test computer output to answer question 13 & 14

INDEPENDENT SAMPLES T-TEST ON ASSERTIVENESS SCORES GROUPED BY GROUP

GROUP N MEAN SD

1 = control 31 73.24 9.00

2 = treatment 31 88.76 8.16

T = 2.97 DF = 60 PROB = < .05

8. Based on the above statistics, the researcher would have to conclude that treatment group is more assertive than the control group?

a. True

b. False

c. Insufficient Information

9. Based on the above, the effect size for the treatment group is 1.7244.

a. True

b. False

10. As relates to the use of a t-test, the term “type II error” refers to:

a. Error that occurs when the null hypothesis is not rejected when there are real differences between the two means that are being compared.

b. reporting findings that are not substantive.

c. Adjusting the level of significance after data has been analyzed.

d. Mistakenly concluding that variables are related when a true relationship does not exist.

11. Given the following statistics, calculate the effect size and its importance for the below t-test for independent samples, where the:

Sample size for treatment group = 51 Sample size for control group = 51

t value = 4.82, df = 100, and p = < .05

Treatment Group Mean = 78 Treatment Group standard deviation = 2.00

Control Group Mean = 75 Control Group standard deviation = 3.24

ES = experimental group mean – control group mean

Control group’s standard deviation

a. .9259 ( moderate - strong effect)

b. .8436 (moderate effect)

c. 4.82 (strong effect)

d. 1.62 (moderate effect)

e. Insufficient information (question can not be answer)

14. Given the same data set for question #23, calculate a 95% confidence interval range for the treatment group mean.

a. 73.18 – 82.82

b. 77.45 – 78.55

c. 74.08 – 81.92

d. 77.12 – 78.88

15. What is the standard error of the mean for the treatment group? SEM= standard deviation (pop.)

Square root of the sample size

a. 3.920

b. .549

c. 4.820

d. .050

e. .280

16. Chi-square determines if there is support for a relationship between variables by comparing:

a. a sample mean with a population mean.

b. the mean of two research samples.

c. observed frequencies and expected frequencies.

d. observed frequencies and degrees of freedom.

17. Chi-square is a test for_____ between variables.

a. cause and effect.

b. correlation.

c. association.

d. all of the above

18. How many degrees of freedom are there in a 2 X 3 crosstabulation table?

a. 6

b. 3

c. 2

d. 1

19. . In chi-square analysis, degrees of freedom provide a way of adjusting for:

a. the size of the research sample.

b. the number of cells that contribute to the chi-square value.

c. the difference between observed and expected frequencies.

d. the difference between nominal and ordinal level measurement.

20. What is a requirement for the use of chi-square?

a. a 2 X 2 cross tabulation table.

b. an expected frequency of 5 or more

c. an observed frequency of 5 or more in at least 80% of the cells.

d. a sample size of at least 100.

APPLIED PROBLEMS: (50 points) Be sure to show your work for credit.

1. You are interested in testing the effectiveness of a newly developed parenting education program. You decide to test the effectiveness of this program by randomly assigning parents to two groups (Group A – the treatment group: Group B – the control group). After 8 weeks, clients from each group are tested using an interval level scale that was specifically developed to assess parenting skills. Test the null hypothesis that there is no difference in parenting scores between group A and group B. The control group standard deviation is 1.134. The treatment group standard deviation is 4.22

Treatment Group Control Group Parenting Knowledge Parenting Knowledge

Maria 18 Esteban 9

Jane 7 Gloria 10

Rene 12 Beth 9

Paul 8 Jorge 7

Chris 17 Tomas 8

Jose 13 Francisca 9

Iris 15 Lisa 7

a. What is the critical t value @ .05 two tail test, i.e., where does null hypothesis rejection region begin?________

b. Decision regarding the Null Hypothesis (circle one):

1. Reject null hypothesis

2. Fail to reject null hypothesis

c. Conclusion related to the effectiveness of the intervention (parenting education classes):

___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

d. Is it appropriate to compute an effect size statistic for this problem? (circle one)

1. Yes

2. No

e, If yes, what is the effect size __________? Explain what this means.

Compute Chi Square for the following cross-tabulation:

Decide whether to reject or fail to reject the null hypothesis that there is no difference between Catholics and Non-Catholics with regards to support or non-support for sex education in elementary schools. Does either group supports sex education in the schools?

Non-

Catholic Catholic Totals

What is the calculated chi-square value? ___________

What are the degrees of freedom _______, and what is the critical chi-square value? ___________

What are the expected frequencies for cells a, b, c, & d)? ________________

Decision: (Circle One) 1. Reject Null Hypothesis

2. Fail to Reject Null Hypothesis Conclusion

Conclusion: (Circle One)

1. There is no difference between Catholics and Non-Catholics with regard to support or non-support of sex education.

2. Catholics are more supportive of sex education than are Non-Catholics.

3. Non-Catholics are more supportive of sex education than are Catholics.

4. Based on the analysis, it isn’t possible to arrive at a conclusion.

Degrees of freedom: 1 Chi-square = 0.105912162162162 For significance at the .05 level, chi-square should be greater than or equal to 3.84. The distribution is not significant. p is less than or equal to 1.

*** The table for Critical Values is found on the next page

Critical Values of t (Level of Significance for a Two-Tailed Test)

df .20 .10 .05 .02 .01

1 3.078 6.314 31.821 31.821 63.657

2 1.886 2.920 4.303 6.965 9.925

3 1.638 2.353 3.182 4.541 5.841

4 1.533 2.132 2.776 3.747 4.604

5 1.476 2.015 2.571 3.365 4.032

6 1.440 1.943 2.447 3.143 3.707

7 1.415 1.895 2.365 2.998 3.499

8 1.397 1.860 2.306 2.896 3.355

9 1.383 1.833 2.262 2.821 3.250

10 1.372 1.812 2.228 2.764 3.169

11 1.363 1.796 2.201 2.718 3.106

12 1.356 1.782 2.179 2.681 3.055

13 1.350 1.771 2.160 2.650 3.012

14 1.345 1.761 2.145 2.624 2.977

15 1.341 1.753 2.131 2.602 2.947

X1 = Treatment Group X2 = Control Group

n1 = sample size in Treatment Group

n2 = sample size in Control Group

SEM = s.d. ÷ √ n in other words:

SEM= s.d (population)

Square root of the sample size

ES = x̄T - x̄C ÷ s.d. control group

df = n -1

df .20 .10 .05 .02 .01 .001

1 1.64 2.71 3.84 5.41 6.64 10.83

2 3.22 4.60 5.99 7.82 9.21 13.82

3 4.64 6.25 7.82 9.84 11.34 16.27

4 5.99 7.78 9.49 11.67 13.28 18.46

5 7.29 9.24 11.07 13.39 15.09 18.46