Question

Some researchers in child development wanted to develop ways to increase the spatial-temporal reasoning of preschool children. This ability, sometimes referred to as thinking in pictures, is important for generating and conceptualizing solutions to multistep problems and is an important aspect of early childhood development. A researcher designed a study to evaluate several methods proposed to accelerate the growth in spatial-temporal reasoning. The researcher wanted to find out, which one, if any, was most effective in increasing development in this area. The 3 methods proposed 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 groups supervised by a trained instructor (instructor). The researcher also included a control group that consisted of children who did not receive any special instruction (control). The effectiveness of the programs was assessed by measuring reasoning before the study and then after the 3 month intervention period. To simplify analysis, the response variable in this study was the difference in the 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 the homogeneity of variance assumption is met. The researcher carried out this test to determine whether the assumption was met for these data. The results are shown below. Based on the results, can the researcher report that it seems reasonable to believe that this assumption has been met?

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

           Df F value    Pr(>F)

group 3 0.0579    0.9816

            96     

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

2. The ANOVA table for this study is shown below. However, some of 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 96 952.30 b                       

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

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

3. The ANOVA result in question 8 is being used to test which one 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 in question 2 (data analysis output: 7.807e-12 ***), what decision would be made about the null hypothesis using a significance level of .01?

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

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

  Posthoc multiple comparisons of means : Tukey HSD

    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 3.3011808 0.69580   

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

control-computer    -4.876 -7.205181 -2.5468192 2.1e-06 ***

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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, the researcher produced the table of descriptive statistics that you see below. Remember, the response variable of interest is the difference in reasoning score (score after intervention period– score before intervention started). Based on the calculation of this variable, a positive value would indicate that the reasoning score was higher after the intervention period and a negative score would indicate that the reasoning score was lower after the intervention period. In general, a higher reasoning score would be considered 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 2 and the information provided, which intervention, if any, would you recommend to early childhood educators as a means of increasing the spatial-temporal reasoning in preschool children?

a. instructor b. piano c. computer d. does not seem to matter

Answer 1

1. For this analysis, the response variable in this study was the difference in the two reasoning scores (score after 3 month intervention period – score at beginning of study). We wanted to see if there is any change in the development of child while using methods such as Piano, video games or games in group for 3 months.

For this, we tested the homogenity of the variance across the 3 months for these 3 techniques and one control group.

Ho: There is homogenity of variance in the data.

H1 : There is no homogeneity of variance in the data.

For this, we did LEVENE test. At 5% level of signifance, p-value = 0.0579 > 0.05 . Therefore, the assumption of homogeneity holds and we fail to reject Ho.

d. Yes, the assumption seems to be met for these data, the test of the null hypothesis of equal variances is not rejected (p>.05).

2. c. 24.437

3. Incomplete question

4. p-value = 7.807e-12 and = 0.01.

Since, p-value < 0.01. We reject Ho.

b. The null hypothesis would be rejected; the obtained p-value is less than the chosen significance level.

5. Ho : The two groups differ signifiantly and = 0.05.

We reject Ho for all those groups where p-value < .

a. computer-piano has p-value 0.6958 > 0.05. We do not reject Ho only in this case. computer-piano group does not differ.

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