Question

Need help to describe the output below. I used the Age group as Factor and Country Music Score as a dependent variable in SPSS and get this output so need to explain the results.

Oneway [DataSetl] Descriptives Rate Music Genre (1=Worst Music ever to 7=Best Music Ever Country 95% Confidence interval for Mean N Mean Std. Deviation Std. Error Lower Bound Upper Bound 11 2.55 .934 282 1.92 3.17 2 7 4.43 2.370 .896 2.24 6.62 3 6 4.67 2.338 .955 2.21 7.12 Total 24 3.63 2.018 412 2.77 4.48 Minimum Maximum 1 4 1 2 7 2 7 1 Country ANOVA Rate Music Genre (1=Worst Music ever to 7=Best Music Ever Sum of Squares df Mean Square Between Groups 23.850 2 11.925 Within Groups 69.775 21 3.323 Total 93.625 23 F Sig 3.589 046 Post Hoc Tests Country 95% Confidence Interval Lower Bound Upper Bound Multiple Comparisons Dependent Variable: Rate Music Genre (1=Worst Music ever to 7=Best Music Ever Tukey HSD (1) Age group (1=18, 19), (J) Age group (1=18, 19). Mean (2=20-22). (3=23 and (2=20-22), (3=23 and Difference (1- over) over) J) Std. Error 2 -1.883 .881 -2.121 .925 2 1.883 .881 - 238 1.014 3 2.121 .925 2 238 1.014 1 Sig .106 .079 -4.10 -4.45 3 1 .34 .21 4.10 2.32 .106 -34 تعا .970 -2.79 1 4.45 .079 970 -21 -2.32 2.79

Answer 1

let

1. μ₁ = Average country score for age group 18-19
2. μ₂ = Average country score for age group 20-22
3. μ_{3 =}Average country score for age group 23 and above

• NULL HYPOTHESIS (H_o) : Average country music score do not differ significantly for different age groups i.e., μ₁ = μ₂ = μ₃
• ALTERNATIVE HYPOTHESIS (H₁) : Average country music score differ significantly for different age groups i.e., μ₁ ≠ μ₂ ≠ μ₃

Now from the ouput we can see that the sig. value for the ANOVA table is less than 0.05 (alpha value = level of significance). So, we may reject our null hypothesis at 0.05 level of significance and conclude that the average country music score differ significantly for different age groups.

POST-HOC test interpretation

Tukey post hoc test is used to test if the two mean differs significantly or not.

If sig. value < 0.05. We conclude that the two means differ significantly.

• Now since the sig. value for factors pairs 1 vs 2, 2 vs 3, 3 vs 1 and 3 vs 2 are greater than 0.05, therefore, the mean of pair of those factors donot differ significantly i.e., μ₁ = μ_{2 ,} μ₂ = μ_{3 ,} μ₁ = μ_{3 ,} μ₂ = μ₃ whereas
• sig. value for factor 1 vs 3 is less than 0.05. Therefore, average of factor 1and factor 2 differ significantly i.e., μ₁ ≠ μ₃