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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