Data Analysis with SPSS : One-way ANOVA

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Data Analysis with SPSSData Analysis with SPSS

One-way ANOVAOne-way ANOVA

Analyze Compare Means One-Way Anova

Example:

We want to examine whether there are significant differences in the monthly salary of employees from different age groups.

Dependent variable : Monthly SalaryIndependent variable : Age Group

MONTHLY SALARY OF RESPONDENT

AGE GROUP OF RESPONDENT

Dependent List

Factor

Press “Post Hoc” Multiple Comparisons Dialog Box

In this example, I have chosen “Scheffe”. Then press “Continue”

Press “OK” to execute

Oneway

ANOVA

MONTHLY SALARY OF RESPONDENT

351.208 2 175.604 132.032 .000

889.778 669 1.330

1240.987 671

Between Groups

Within Groups

Total

Sum ofSquares df Mean Square F Sig.

F = 132.032, Sig. = .000

Shows that the mean salary of the three age groups are significantly different

We do not know which group means are different, post hoc test will indicate this

Post Hoc Tests

Multiple Comparisons

Dependent Variable: MONTHLY SALARY OF RESPONDENT

Scheffe

-1.081* .111 .000 -1.35 -.81

-2.003* .123 .000 -2.30 -1.70

1.081* .111 .000 .81 1.35

-.922* .105 .000 -1.18 -.66

2.003* .123 .000 1.70 2.30

.922* .105 .000 .66 1.18

(J) AGE GROUP OFRESPONDENT26 - 35 YEARS

36 YEARS AND ABOVE

25 YEARS AND BELOW

36 YEARS AND ABOVE

25 YEARS AND BELOW

26 - 35 YEARS

(I) AGE GROUP OFRESPONDENT25 YEARS AND BELOW

26 - 35 YEARS

36 YEARS AND ABOVE

MeanDifference

(I-J) Std. Error Sig. Lower Bound Upper Bound

95% Confidence Interval

The mean difference is significant at the .05 level.*.

Scheffe Multiple Comparisons test shows that all the three group means are significantly different from one another, sig. (or p) ≤ 0.001

Lets look at two other examples

ANOVA

impgdevt

.376 2 .188 .370 .691

339.527 669 .508

339.902 671

Between Groups

Within Groups

Total

Sum ofSquares df Mean Square F Sig.

ANOVA to test whether there is/are significant difference(s) in the means of “importance of growth and development” between employees of different age groups

F = 0.370, p = 0.691

p >0.05, so there is no significant difference between the means of the three age groups for the importance of “growth and development”

Example 1

Post Hoc Tests

Multiple Comparisons

Dependent Variable: impgdevt

Scheffe

-.01747 .06887 .968 -.1864 .1515

.03839 .07613 .881 -.1484 .2251

.01747 .06887 .968 -.1515 .1864

.05586 .06507 .692 -.1038 .2155

-.03839 .07613 .881 -.2251 .1484

-.05586 .06507 .692 -.2155 .1038

(J) AGE GROUP OFRESPONDENT26 - 35 YEARS

36 YEARS AND ABOVE

25 YEARS AND BELOW

36 YEARS AND ABOVE

25 YEARS AND BELOW

26 - 35 YEARS

(I) AGE GROUP OFRESPONDENT25 YEARS AND BELOW

26 - 35 YEARS

36 YEARS AND ABOVE

MeanDifference

(I-J) Std. Error Sig. Lower Bound Upper Bound

95% Confidence Interval

All the significant levels are more than 0.05, so there is no difference in the means of the groups

Example 2

ANOVA

penvr

3.975 2 1.987 3.911 .020

339.927 669 .508

343.902 671

Between Groups

Within Groups

Total

Sum ofSquares df Mean Square F Sig.

ANOVA to test whether there is/are significant difference(s) in the means of “importance of safe work environment (penvr)” between employees of different age groups

F = 3.911, p = 0.02

p = 0.02, (i.e. ≤ 0.05), so there is significant difference between the means

Multiple Comparisons

Dependent Variable: penvr

Scheffe

.08662 .06891 .454 -.0824 .2557

.20965* .07617 .023 .0228 .3965

-.08662 .06891 .454 -.2557 .0824

.12303 .06511 .169 -.0367 .2828

-.20965* .07617 .023 -.3965 -.0228

-.12303 .06511 .169 -.2828 .0367

(J) AGE GROUP OFRESPONDENT26 - 35 YEARS

36 YEARS AND ABOVE

25 YEARS AND BELOW

36 YEARS AND ABOVE

25 YEARS AND BELOW

26 - 35 YEARS

(I) AGE GROUP OFRESPONDENT25 YEARS AND BELOW

26 - 35 YEARS

36 YEARS AND ABOVE

MeanDifference

(I-J) Std. Error Sig. Lower Bound Upper Bound

95% Confidence Interval

The mean difference is significant at the .05 level.*.

Post Hoc Tests

Scheffe test shows that there is significant difference between a pair of means: “25 YEARS AND BELOW” and “36 YEARS AND ABOVE”, p = 0.023 (≤0.05)

Thank You

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