Methods and Analysis of Quantitative Research Essay

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Name of Database: R7031education.sav

Name of Assignment File: R7031education.sav assignments.docx

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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 hypothesis

There 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 collection

In 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-value

Using 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) = 21.193

The p-value (sig. 2-tailed) is p < .001.

Table 1

                                                                                                        Descriptives

Safety of Children

N
Mean
Std. Deviation
Std. Error
95% Confidence Interval for Mean
Minimum
Maximum
Lower BoundUpper Bound

One Child
18

3.056

1.3492

.3180

2.385

3.726

1.0

5.0

Two Children
18
4.389
1.2433
.2930
3.771
5.007
1.0
6.0
Three or More Children
14
5.929
1.0716
.2864
5.310
6.547
4.0
7.0
Total
50
4.340
1.6734
.2367
3.864
4.816
1.0
7.0

Table 2

                                                  Test of Homogeneity of Variances

Safety of Children

Levene Statistic
df1
df2
Sig.
.422
2
47
.658

Table 3

                                                                              ANOVA

Safety of Children

Sum of Squares
df
Mean Square
F
Sig.
Between Groups
65.069
2
32.535
21.193
.000
Within Groups
72.151
47
1.535

Total
137.220
49

Step 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 assessment

Risk 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 result

The 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 hypothesis

The null and alternate hypotheses are

Step 2: Level of significance, ?

The level of significance, ? selected is .05.

Step 3: The data collection

In 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-value

Comparing 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 Comparisons

Dependent Variable: Safety of Children

Tukey HSD

(I) Number of Children in School
(J) Number of Children in School
Mean Difference (I-J)
Std. Error
Sig.
95% Confidence Interval
Lower BoundUpper Bound
One Child
Two Children
-1.3333(*)
.4130
.006
-2.333
-.334
Three or More Children-2.8730(*)
.4415
.000
-3.942
-1.804
Two Children
One Child
1.3333(*)
.4130
.006
.334
2.333
Three or More Children-1.5397(*)
.4415
.003
-2.608
-.471
Three or More Children
One Child
2.8730(*)
.4415
.000
1.804
3.942
Two Children1.5397(*)
.4415
.003
.471
2.608
*  The mean difference is significant at the .05 level.

Step 5: Retain or reject the null hypothesis.

 For :

One Child vs. Two Children
Reject the null hypothesis
One Child vs. Three or More Children
Reject the null hypothesis
Two Children vs. Three or More Children
Reject the null hypothesis

Step 6: The Risk of Type I and Type II Error assessment

Tukey’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 result

The 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 hypothesis

For 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 collection

In 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-value

As 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 Statistics

Dependent Variable: Overall Satisfaction in January

Type of Student
Receives Title 1 Free Lunch
Mean
Std. Deviation
N
New Student
No
4.3571
1.27745
14
Yes3.6154
1.26085
13
Total4.0000
1.30089
27
Returning Student
No
2.5000
1.16775
12
Yes3.7273
1.19087
11
Total3.0870
1.31125
23
Total
No
3.5000
1.52971
26
Yes3.6667
1.20386
24
Total3.5800
1.37158
50

Table 6

               Levene’s Test of Equality of Error Variances(a)

Dependent Variable: Overall Satisfaction in January

F
df1
df2
Sig.
.092
3
46
.964
Tests the null hypothesis that the error variance of the dependent variable is equal across groups.

a  Design: Intercept+Newstudent+Title1+Newstudent * Title1

Table 7

                                                                      Tests of Between-Subjects Effects

Dependent Variable: Overall Satisfaction in January

Source
Type III Sum of Squares
df
Mean Square
F
Sig.
Partial Eta Squared
Corrected Model
22.707(a)
3
7.569
5.012
.004
.246
Intercept
625.040
1
625.040
413.856
.000
.900
Newstudent
9.442
1
9.442
6.252
.016
.120
Title1
.731
1
.731
.484
.490
.010
Newstudent * Title1
12.018
1
12.018
7.958
.007
.147
Error
69.473
46
1.510

Total
733.000
50

Corrected Total
92.180
49

a  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 assessment

Risk 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 result

The 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 hypothesis

Hypothesis 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 = ?returning

H1: Mnew ? Mreturning

Step 2: Level of significance, ?

The level of significance, ? selected is .05.

Step 3: The data collection

In 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 Statistics

Dependent Variable: Overall Satisfaction in June

Type of Student
Mean
Std. Deviation
N
New Student
5.111
.9337
27
Returning Student
4.217
.7359
23
Total
4.700
.9530
50

Table 9

        Levene’s Test of Equality of Error Variances(a)

Dependent Variable: Overall Satisfaction in June

F
df1
df2
Sig.
1.729
1
48
.195
Tests the null hypothesis that the error variance of the dependent variable is equal across groups.

a  Design: Intercept+Osatjan+Newstudent

Table 10

                                                                   Tests of Between-Subjects Effects

Dependent Variable: Overall Satisfaction in June

Source
Type III Sum of Squares
df
Mean Square
F
Sig.
Partial Eta Squared
Corrected Model
15.089(a)
2
7.544
12.056
.000
.339
Intercept
82.041
1
82.041
131.105
.000
.736
Osatjan
5.169
1
5.169
8.260
.006
.149
Newstudent
4.864
1
4.864
7.774
.008
.142
Error
29.411
47
.626

Total
1149.000
50

Corrected Total
44.500
49

a  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 assessment

Risk 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 result

The 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.

CONCLUSION

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. 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 University

Online 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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