Read the Following Scenario and Use the Data Contained on Page 3 to Answer the Questions

EPS 625 – INTERMEDIATE STATISTICS PRACTICE t TEST – NO. 2 – KEY

Dr. Kureous, a teacher at George Junior High, now wants to determine if there is a significant difference between the 6th grade boys and girls in his class on their mathematics exam. He does not have a prediction as to whether the boys or the girls will be better – he simply wants to see if they differ significantly on the math exam.

1. State (using symbols or words) the null hypothesis and the alternative hypothesis for his mathematics Independent-Samples t Test.

H0: 1 = 2 or 1 – 2 = 0

There is no (statistically significant) difference between the boys’ mathematics

exam mean (1) and the girls’ mathematics exam mean (2).  Could also make reference to the hypothesized mean difference = 0.

Ha: 1  2 or 1 – 2  0

There is a (statistically significant) difference between the boys’ mathematics

exam mean (1) and the girls’ mathematics exam mean (2).  Could also make reference to the hypothesized mean difference  0.

2. Using an alpha () level of .05. 2.a. Has Dr. Kureous met the assumption of homogeneity of variance for the Math Exam t test?

YES or NO (circle your answer selection)

2.b. Justify/explain your answer – that is, how did you come to your conclusion? (Hint: look at the SPSS printout).

The Levene’s Test for Equality of Variances showed: F = 8.033, p = .008 Compared to  = .05, p <  (.008 < .05) – reject the null hypothesis of no difference. Therefore it is significant and the assumption is not met – that is, equal variances are not assumed.

That is: p (.008) < α (.05) 3. After computing the test statistic (t-test) – complete the following tables for the Math Exam Independent-Samples t-Test (Do not round – use the values from SPSS):

Standard Standard Error Gender N Mean Deviation of the Mean Boys 15 88.67 6.532 1.687

Girls 15 80.40 12.642 3.264

95% CI of the Sig. Mean t df Difference (2-tailed) Difference Lower Upper 2.250 20.977 .035 8.267 .625 15.908

4. Using an alpha () level of .05, interpret the results from the Math Exam t test: 4.a. Did you reject the null hypothesis in favor of the alternative hypothesis (indicated in question 1)?

4.b. Justify your answer, that is, how did you come to your conclusion? DO NOT make reference to the t critical value or the confidence intervals. If applicable (or indicate otherwise and why), calculate the Effect Size – be sure to show your work.

t (20.977) = 2.250, p = .035. Compared to  = .05, p <  (.035 < .05) – which is significant, therefore the null hypothesis of no difference is rejected. That is: p (.035) < α (.05).

N  N 15 15 30 d  t 1 2  2.250  2.250  2.250 .133333  2.250(.365148) N1 N 2 (15)(15) 225 d = .8215735 = .82

4.c. Briefly discuss your findings (i.e., what do the results indicate or mean). Be sure to make reference to the group means and include all applicable values. Also include the statistical string (values and symbols).

The boys (M = 88.67, SD = 6.53) performed significantly higher than the girls (M = 80.40, SD = 12.64) on the mathematics test for this sample at the .05 alpha level, with an effect size of nearly one (d = .82) standard deviation.

t(20.98) = 2.25, p < .05, d = .82

PRACTICE t TEST – NO. 2 – KEY PAGE – 2 Dr. Stats would now like to see if there is a difference between Exam_2 and the Final for her entire class. She does not have a prediction as to whether the performance will be higher or lower for the two exam comparison – she simply wants to determine if the class performed significantly different on the two test occasions.

5. Using an alpha () level of .05, interpret the results from the Exam_2 / Final t test:

5.a. Using the Table of Critical Values of Student’s t Distribution – what would the t critical value (tcv) be for the Exam_2 / Final Dependent-Samples t Test?

With a two-tailed test using an alpha level of .05 and df = 44 (we use 40

degrees of freedom), we find tcv = 2.021

5.b. Did you reject the null hypothesis in favor of the alternative hypothesis?

5.c. Justify your answer, that is, how did you come to your conclusion? You may make reference to any of the three methods of determination. Be sure to include all applicable values for the method that you choose. If applicable (or indicate otherwise and why), calculate the Effect Size – be sure to show your work.

1) Comparing the Sig. (probability) = .861 to the a priori alpha level,  = .05  p >  – therefore, we retain the null hypothesis of no difference.

2) Comparing tobt = |.176| to the tcv = |2.021|.

 tobt < tcv – therefore, we retain the null hypothesis of no difference. 3) Examining the confidence interval (Lower = -4.638, Upper = 5.527) to see if zero (0) is contained within it’s limits.

 The confidence interval CI95 does contain zero – therefore, we retain the null hypothesis of no difference.

No Effect Size calculation is needed since there was a non-significant finding.

5.d. Briefly discuss your findings (i.e., what do the results indicate or mean). Be sure to make reference to the group means and include all applicable values. Also include the statistical string (values and symbols).

While the mean for Exam_2 (M = 81.33, SD = 13.92) was higher than the mean for the Final (M = 80.89, SD = 13.28), it was not significantly different. The mean difference (.444) was not significantly different from zero (0) at the . 05 alpha level. Any difference can be attributed to sampling error (sampling fluctuation or group characteristics) for this sample.

t(44) = .18, p > .05 or t(44) = .176, p = .861

PRACTICE t TEST – NO. 2 – KEY PAGE – 3