Some researchers in child development wanted to develop ways toincrease the spatial-temporal reasoning of preschool children. Thisability, sometimes referred to as thinking in pictures, isimportant for generating and conceptualizing solutions to multistepproblems and is an important aspect of early childhood development.A researcher designed a study to evaluate several methods proposedto accelerate the growth in spatial-temporal reasoning. Theresearcher wanted to find out, which one, if any, was mosteffective in increasing development in this area. The 3 methodsproposed included: taking piano lessons for 3 months (piano),playing specially developed computer video games for 3 months(computer), and playing specially developed games in small groupssupervised by a trained instructor (instructor). The researcheralso included a control group that consisted of children who didnot receive any special instruction (control). The effectiveness ofthe programs was assessed by measuring reasoning before the studyand then after the 3 month intervention period. To simplifyanalysis, the response variable in this study was the difference inthe two reasoning scores (score after 3 month intervention period –score at beginning of study).

1. Levene’s test is a common test used to assess whether thehomogeneity of variance assumption is met. The researcher carriedout this test to determine whether the assumption was met for thesedata. The results are shown below. Based on the results, can theresearcher report that it seems reasonable to believe that thisassumption has been met?

Levene's Test for Homogeneity of Variance (center = \"mean\")

           DfF value    Pr(>F)

group 3 0.0579    0.9816

           96     

a. No, the assumption is not met for these data, the test of thenull hypothesis of equal variances is rejected (p<.05). b. No,the assumption is not met for these data, the test of the nullhypothesis of equal variances is not rejected (p>.05). c. Yes,the assumption seems to be met for these data, the test of the nullhypothesis of equal variances is rejected (p<.05). d. Yes, theassumption seems to be met for these data, the test of the nullhypothesis of equal variances is not rejected (p>.05).

2. The ANOVA table for this study is shown below. However, someof the values from the table are missing. What is the F value(c)?

Analysis of Variance Table

Response: reasoningdiff

Df Sum Sq Mean Sq F value Pr(>F)   

method 3 727.24 a c 7.807e-12 ***

Residuals 96 952.30b                       

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’1

a. 242.413 b. Cannot be determined from the informationprovided. c. 24.437 d. 9.92

3. The ANOVA result in question 8 is being used to test whichone of the following null hypotheses?

a. H_o:   b. H_o:   c.H_o:   d. H_o:

4. Consider the p-value that is in the output presented inquestion 2 (data analysis output: 7.807e-12 ***), what decisionwould be made about the null hypothesis using a significance levelof .01?

a. The null hypothesis would not be rejected; the obtainedp-value is greater than the chosen significance level. b. The nullhypothesis would be rejected; the obtained p-value is less than thechosen significance level. c. The null hypothesis would berejected; the obtained p-value is greater than the chosensignificance level. d. The null hypothesis would not be rejected;the obtained p-value is less than the chosen significancelevel.

5. Assuming it was appropriate to conduct post hoc tests, thefollowing results were obtained for the child development study.Based on these results and using a significance level of .05, whichgroups do not differ?

  Posthoc multiple comparisons of means : TukeyHSD

    95% family-wise confidence level

$method

diff lwr.ci upr.ci pval   

piano-instructor -3.608 -5.937181 -1.2788192 0.00059 ***

computer-instructor -2.636 -4.965181 -0.3068192 0.01998 *

control-instructor -7.512 -9.841181 -5.1828192 4.3e-10 ***

computer-piano 0.972 -1.357181 3.30118080.69580   

control-piano -3.904 -6.233181 -1.5748192 0.00017 ***

control-computer    -4.876 -7.205181 -2.54681922.1e-06 ***

---

Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' '1

a. computer-piano b. control-computer c. piano-instructor d.computer-instructor

6. In reporting the results of the child development study, theresearcher produced the table of descriptive statistics that yousee below. Remember, the response variable of interest is thedifference in reasoning score (score after intervention period–score before intervention started). Based on the calculation ofthis variable, a positive value would indicate that the reasoningscore was higher after the intervention period and a negative scorewould indicate that the reasoning score was lower after theintervention period. In general, a higher reasoning score would beconsidered a better outcome.

method count mean sd

1 instructor    25 7.4 3.37   

2 piano 25 3.79 2.75    

3 computer 25 4.76 3.14    

4 control 25 -0.112 3.31   

Based on the descriptive table, the results shown in question 2and the information provided, which intervention, if any, would yourecommend to early childhood educators as a means of increasing thespatial-temporal reasoning in preschool children?

a. instructor b. piano c. computer d. does not seem tomatter