Methods and Analysis of Quantitative Research Essay

————————————————————————————————————Name of Database: R7031education.

savName of Assignment File: R7031education.sav assignments.docx————————————————————————————————————Question 1: Are there differences in parent satisfaction with the safety of their children between those who have one child, two children, or three or more in daycare?Step 1: The null and alternate hypothesisThere are not differences in parent satisfaction with the safety of their children between those who have one child, two children, or three or more in daycare.There are differences in parent satisfaction with the safety of their children between those who have one child, two children, or three or more in daycare.Step 2: Level of significance, ?The level of significance, ? selected is .05.

Step 3: The data collectionIn January 08, One hundred elementary school parents of a county school district were surveyed regarding their satisfaction with a free after-school daycare program. The parents filled out the survey on after the end of the fall semester (in January 08) (database R7031education.sav).

Step 4: The test statistic and p-valueUsing SPSS One-Way ANOVA,            Visual comparison of three sample means, 3.06 vs. 4.39 vs. 5.93 suggest that parent satisfaction with the safety of their children in daycare is higher for three or more, and two children as compared to one child.Levene’s test: F(2, 47)=.422, p = .

658. The Sig value is > .05, so the null hypothesis is accepted. The variances are equal. The ANOVA assumption of homogeneity of variance was not violated.The test statistic is F (2, 47) = 21.193The p-value (sig. 2-tailed) is p < .

001.Table 1                                                                                                        DescriptivesSafety of ChildrenNMeanStd. DeviationStd.

Error95% Confidence Interval for MeanMinimumMaximumLower BoundUpper BoundOne Child183.0561.3492.31802.

3853.7261.05.0Two Children184.3891.

2433.29303.7715.

0071.06.0Three or More Children145.9291.0716.28645.

3106.5474.07.0Total504.3401.6734.23673.

8644.8161.07.

0Table 2                                                  Test of Homogeneity of VariancesSafety of ChildrenLevene Statisticdf1df2Sig..422247.658Table 3                                                                              ANOVASafety of ChildrenSum of SquaresdfMean SquareFSig.Between Groups65.069232.

53521.193.000Within Groups72.151471.535Total137.22049Step 5: Retain or reject the null hypothesis. Since, F (2, 47) = 21.193, p < .

001. Therefore, reject the null hypothesis.Step 6: The Risk of Type I and Type II Error assessmentRisk of Type I error (the risk of incorrectly rejecting the null). The calculated p value is less than 0.001, well below alpha, so the risk of Type I error is very low.Effect size,Eta, =The results indicate a very strong effect.Step 7: APA format resultThe results of the analysis are, F (2, 47) = 21.193, p < .

001. Since the p-value is less than .05, we reject the null hypothesis. The difference is statistically significant. Therefore, there are differences in parent satisfaction with the safety of their children between those who have one child, two children, or three or more in daycare.Post Hoc Comparison: How are parent satisfactions with the safety of their children between those who have one child, two children, or three or more in daycare is different from one another?Step 1: The null and alternate hypothesisThe null and alternate hypotheses areStep 2: Level of significance, ?The level of significance, ? selected is .

05.Step 3: The data collectionIn January 08, One hundred elementary school parents of a county school district were surveyed regarding their satisfaction with a free after-school daycare program. The parents filled out the survey on after the end of the fall semester (in January 08) (database R7031education.sav).Step 4: The test statistic and p-valueComparing One Child vs. Two Children shows a mean difference of -1.33, p = .

006. Therefore, there is statistically significant difference.Comparing One Child vs.

Three or More Children shows a mean difference of -2.87, p < .001. Therefore, there is statistically significant difference.Comparing Two Children vs.

Three or More Children shows a mean difference of -1.54, p = .003. Therefore, there is statistically significant difference.Table 4                                                                                      Multiple ComparisonsDependent Variable: Safety of ChildrenTukey HSD(I) Number of Children in School(J) Number of Children in SchoolMean Difference (I-J)Std.

ErrorSig.95% Confidence IntervalLower BoundUpper BoundOne ChildTwo Children-1.3333(*).4130.006-2.

333-.334Three or More Children-2.8730(*).4415.

000-3.942-1.804Two ChildrenOne Child1.3333(*).4130.006.3342.

333Three or More Children-1.5397(*).4415.003-2.

608-.471Three or More ChildrenOne Child2.8730(*).

4415.0001.8043.942Two Children1.5397(*).

4415.003.4712.

608*  The mean difference is significant at the .05 level.Step 5: Retain or reject the null hypothesis.

 For :One Child vs. Two ChildrenReject the null hypothesisOne Child vs. Three or More ChildrenReject the null hypothesisTwo Children vs. Three or More ChildrenReject the null hypothesisStep 6: The Risk of Type I and Type II Error assessmentTukey’s HSD (Honestly Significant Differences) is a conservative test and by choosing it, the risk of Type I error is minimized.Step 7: APA format resultThe hypothesis that there are differences in parent satisfaction with the safety of their children between those who have one child, two children, or three or more in daycare was supported, F (2, 47) = 21.193, p ;  .001, the results of the Tukey’s post-hoc comparisons indicated that all three means differed significantly from one another. One Child and Two Children differed significantly, mean difference = -1.

33, p = .006; One Child and Three or More Children differed significantly, mean difference = -2.87, p ; .001; Two Children and Three or More Children differed significantly, mean difference = -1.54, p = .003.Question 2: Did type of student (new vs.

returning) and the Free Lunch program affect overall parent satisfaction in January?Step 1: The null and alternate hypothesisFor the research question, there will be three hypotheses.Are there any differences between the types of student?                   (Main Effect 1)Are there any differences between the free lunch programs?                   (Main effect 2)Is there a combined relationship of the types of student and the free lunch program that has effect on overall parent satisfaction in January?                                         (Interaction)Step 2: Level of significance, ?The level of significance, ? selected is .05.Step 3: The data collectionIn January 08, One hundred elementary school parents of a county school district were surveyed regarding their satisfaction with a free after-school daycare program. The parents filled out the survey on after the end of the fall semester (in January 08) (database R7031education.sav).

Step 4: The test statistic and p-valueAs shown in table 5, examination of the means reveals that there are differences between the free lunch programs for each type of student.Levene’s test: F(3, 46)=.092, p = .964. The Sig. value is ; .

05, so the null hypothesis is retained. The variances are equal; therefore, the assumptions are not violated.The three most important rows to look at are those highlighted (table 7).The row Newstudent indicates the effect of the type of student, F (1, 46) = 6.25, p = .016.

There is a main effect for type of student.The row Title 1 indicates the effect of the free lunch programs, F(1, 46) = .48, p =.490. There is no main effect for the free lunch programs.

The row Newstudent * Title1 is the interaction effect, F(1, 46) = 7.96, p = .007. The interaction is significant.Table 5                                                          Descriptive StatisticsDependent Variable: Overall Satisfaction in JanuaryType of StudentReceives Title 1 Free LunchMeanStd.

DeviationNNew StudentNo4.35711.2774514Yes3.

61541.2608513Total4.00001.3008927Returning StudentNo2.50001.1677512Yes3.72731.

1908711Total3.08701.3112523TotalNo3.50001.5297126Yes3.66671.

2038624Total3.58001.3715850Table 6               Levene’s Test of Equality of Error Variances(a)Dependent Variable: Overall Satisfaction in JanuaryFdf1df2Sig..

092346.964Tests the null hypothesis that the error variance of the dependent variable is equal across groups.a  Design: Intercept+Newstudent+Title1+Newstudent * Title1Table 7                                                                      Tests of Between-Subjects EffectsDependent Variable: Overall Satisfaction in JanuarySourceType III Sum of SquaresdfMean SquareFSig.Partial Eta SquaredCorrected Model22.707(a)37.5695.012.004.

246Intercept625.0401625.040413.

856.000.900Newstudent9.44219.4426.252.

016.120Title1.7311.731.

484.490.010Newstudent * Title112.018112.0187.958.007.147Error69.

473461.510Total733.00050Corrected Total92.18049a  R Squared = .246 (Adjusted R Squared = .197)Step 5: Retain or reject the null hypothesis.

Reading from the three rows:Main Effect 1: Type of student (Newstudent) is statistically significant, F (1, 46) = 6.25, p = .016. Therefore, reject the null hypothesis.Main Effect 2: The free lunch programs (Title 1) is not statistically significant, F(1, 46) = .48, p =.490.

Therefore, retain the null hypothesis.Interaction: There is a significant interaction— overall parent satisfaction in January depends on The free lunch programs in combination with the type of student, F(1, 46) = 7.96, p = .007.

Therefore, reject the null hypothesis.Step 6: The Risk of Type I and Type II Error assessmentRisk of Type I error: ANOVA—by its very design—reduces the risk of Type I error.Risk of Type II error: The effect size are (the statistic eta) .12, .01, and .15, respectively. The interaction term has the largest effect size, although it is rather small.For type of student, which was significant, eta = .

12. This indicates a small effect. For the free lunch programs, which was not significant, eta = .01, indicating no effect of this on overall parent satisfaction in January. The interaction between type of student and the free lunch programs was statistically significant and showed a small effect, eta = .15.

Step 7: APA format resultThe results indicate a significant main effect for Type of student, F(1, 46) = 6.25, p = .016. A significant effect was also found for the interaction term, F(1, 46) = 7.

96, p = .007. The free lunch programs by itself was not a significant variable, F(1, 46) = .48, p =.490.

 As a whole, type of student (new vs. returning) and the Free Lunch program affect overall parent satisfaction in January.Question 3: Is there an improvement in parent satisfaction from January to June? And, are new students more satisfied than returning students?Here there are two research questions to evaluate.

·         Does the covariate have an impact?·         Is there a difference between the groups after the effect of the covariate is removed?Step 1: The null and alternate hypothesisHypothesis 1: Does the covariate have an impact on the dependent variable?H0 : There is no relationship between the covariate and the dependent variable.H1: There is a relationship between the covariate and the dependent variable.Hypothesis 2: Is there a difference between groups once the effect of the covariate is removed?H0: ?new = ?returningH1: Mnew ? MreturningStep 2: Level of significance, ?The level of significance, ? selected is .05.Step 3: The data collectionIn January 08, One hundred elementary school parents of a county school district were surveyed regarding their satisfaction with a free after-school daycare program.

The parents filled out the survey on after the end of the fall semester (in January 08). In June 08, the parents were telephoned and re-surveyed and were asked to rate their overall satisfaction again (database R7031education.sav).

Step 4: The test statistic and p-value            The means look somewhat different (table 8). However, we see if in the covariate analysis whether this is due to parent satisfaction in January.Levene’s test: F(1, 48)=.1.73, p = .

195. The Sig. value is > .05, so the null hypothesis is retained. The variances are equal; therefore, the assumptions are not violated.Table 8                                          Descriptive StatisticsDependent Variable: Overall Satisfaction in JuneType of StudentMeanStd.

DeviationNNew Student5.111.933727Returning Student4.

217.735923Total4.700.953050Table 9        Levene’s Test of Equality of Error Variances(a)Dependent Variable: Overall Satisfaction in JuneFdf1df2Sig.1.729148.195Tests the null hypothesis that the error variance of the dependent variable is equal across groups.

a  Design: Intercept+Osatjan+NewstudentTable 10                                                                   Tests of Between-Subjects EffectsDependent Variable: Overall Satisfaction in JuneSourceType III Sum of SquaresdfMean SquareFSig.Partial Eta SquaredCorrected Model15.089(a)27.54412.

056.000.339Intercept82.041182.041131.105.

000.736Osatjan5.16915.

1698.260.006.

149Newstudent4.86414.8647.774.008.

142Error29.41147.626Total1149.00050Corrected Total44.50049a  R Squared = .

339 (Adjusted R Squared = .311)The first row (yellow) “Parent satisfaction in January (Osatjan)” is the covariate—what was parent satisfaction was in January, and how much impact it has on parent satisfaction in June. The results indicate F(1, 47) = 8.260, p =.006. This is a very strong effect.The second row (green) “Type of student (Newstudent)” is the main effect— are new students more satisfied than returning students? The results indicate F(1, 47) = 7.774, p = .

008. There is difference here.Step 5: Retain or reject the null hypothesis.

Hypothesis 1. The results indicate F(1, 47) = 8.260, p =.006. Therefore, reject the null hypothesis.Hypothesis 2. The results indicate F(1, 47) = 7.774, p = .

008. Therefore, reject the null hypothesis.Step 6: The Risk of Type I and Type II Error assessmentRisk of Type I error.

ANOVA—by its very design—reduces the risk of Type I error.The effect size of the covariate and the main effect are 0.149, and 0.

142, respectively. Therefore, there is small effect for both the covariate and the main effect.Step 7: APA format resultThe results of the analysis indicate that the covariate (parent satisfaction in January) accounts for the differences in parent satisfaction in June F(1, 47) = 8.

260, p =.006. The results also indicate that, if we take into account parent satisfaction in January, that there is difference between satisfaction for new student and returning student F(1, 47) = 7.774, p = .008.In other words, there is an improvement in parent satisfaction from January to June and new students are more satisfied as compared to returning students.CONCLUSIONThere are differences in parent satisfaction with the safety of their children between those who have one child, two children, or three or more in daycare.

Further, parent satisfaction with the safety of their children for one child, two children, or three or more in daycare differed significantly from one another.The type of student affected overall parent satisfaction in January and the Free Lunch program did not affect overall parent satisfaction in January. However, there was a significant effect of interaction of type of student and the Free Lunch program. As a whole, type of student (new vs.

returning) and the Free Lunch program affect overall parent satisfaction in January.There is an improvement in parent satisfaction from January to June and new students are more satisfied as compared to returning students.Reference:Methods and Analysis of Quantitative Research (January, 2009).  Argosy UniversityOnline Course. E7031. Instructor: Michael Marrapodi.  Retrieved February 18, 2009, from http://www.myeclassonline.com/ec/crs/default.learn?CourseID=3282843;CPURL=www.myeclassonline.com;Survey=1;47=4157986;ClientNodeID=404511;coursenav=0;bhcp=1

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