Question

A researcher measured the number of hours of community service performed during an average week in a sample for three adolescent and young adult age groups. The results are shown below.  

14 year olds                17 year olds                21 year olds

            5                                              8                                              3         

         6                                                12                                            5

          3                                                7                                              7

            5                                              4                                              4                

            

Conduct a one-way ANOVA to determine if there are any differences between the groups on community service. The test is 2 –tailed, with an alpha of .05, and the critical F is 4.26. Make sure answers are clearly labeled and well-organized.

Step 1 (1 pts) Restate the question as an alternative hypothesis and a null hypothesis about the populations

Step 2 (1 pts) Determine the characteristics of the comparison distribution.

Step 3 (1 pt) Find the threshold (critical) value on the comparison distribution

Step 4 (3 pts) Determine your sample’s score on the comparison distribution.

Answer 1

solution: from the given data,

• Mean of 1970’s data:
• Mean of 1980’s data:
• Mean of 1990’s data:

In this example, the hypotheses are:

Null hypothesis There is no significant difference in the average number of hours of community service in three age group.
Alternative hypothesis At least one the average number of hours of community service is different among the three age group.

Null hypothesis H₀: μ₁ = μ₂ = μ₃

Alternative hypothesis H₁: The means are not all equal.

Significance level α = 0.05

Equal variances were assumed for the analysis.

Factor Information

Factor Levels Values
Age group 3   14 year old, 17 year old, 21 year old

The test statistic for testing H₀: μ₁ = μ₂ = μ3 is:

where

• = individual observation,

• = sample mean of the j^th treatment (or group),
• = overall sample mean,
• k = the number of treatments or independent comparison groups, and
• N = total number of observations or total sample size.

In order to determine the critical value of F we need degrees of freedom, df₁=k-1 and df₂=N-k. In this example, df₁=k-1=3-1=2 and df₂=N-k=12-3=9.

The critical value is

and the decision rule is as follows: Reject H₀ if F > 4.256.

Analysis of Variance

Source	DF	Adj SS	Adj MS	F-Value	P-Value
Factor	2	24	12	2.34	0.15195
Error	9	46.25	5.139
Total	11	70.25

Conclusion: P-value=0.15195

We fail to reject H₀ because F-value=2.34 < F-critical value= 4.256 and also P-value=0.15195> level of significance=0.05.

We have statistically no significant evidence at α=0.05

Hence, There is no significant difference in the average number of hours of community service in three age group. please give me like, thank you