To compute the ANOVA (Analysis of Variance) for the scores obtained by early, middle, and late adolescents on the adjustment scale, we need to follow several steps.
Let’s go through them:
Step 1: State the hypotheses:
– Null hypothesis (H0): There is no significant difference between the means of the scores of early, middle, and late adolescents on the adjustment scale.
– Alternative hypothesis (Ha): There is a significant difference between the means of the scores of early, middle, and late adolescents on the adjustment scale.
Step 2: Calculate the necessary statistics:
We will compute the sum of squares (SS), degrees of freedom (df), mean squares (MS), and the F-statistic.
First, let’s calculate the sum of squares:
– Total Sum of Squares (SST): This measures the total variability in the data.
– Treatment Sum of Squares (SSTR): This measures the variability between the group means.
– Error Sum of Squares (SSE): This measures the variability within each group.
To calculate these sums of squares, we need the following formulas:
SST = SSTR + SSE
SSTR = Σ(ni * (mean_i – grand_mean)^2)
SSE = Σ((xi – mean_i)^2)
Where:
– ni is the number of observations in the i-th group.
– mean_i is the mean of the i-th group.
– grand_mean is the mean of all observations.
– xi is each individual observation.
Given the scores, let’s calculate the necessary statistics:
Early Adolescents:
N1 = 10, mean1 = (2+3+2+3+4+2+4+5+2+2)/10 = 27/10 = 2.7
Middle Adolescents:
N2 = 10, mean2 = (3+2+4+2+4+3+2+4+3+4)/10 = 31/10 = 3.1
Late Adolescents:
N3 = 10, mean3 = (7+6+5+4+2+3+4+4+3+2)/10 = 40/10 = 4.0
Now let’s calculate the sums of squares:
Grand_mean = (2.7 + 3.1 + 4.0)/3 = 9.8/3 = 3.267
SSTR = (10 * (2.7 – 3.267)^2) + (10 * (3.1 – 3.267)^2) + (10 * (4.0 – 3.267)^2)
= 8.967 + 0.477 + 4.864
= 14.308
SSE = ((2 – 2.7)^2) + ((3 – 2.7)^2) + … + ((4 – 4.0)^2)
= 0.49 + 0.09 + … + 0.16
= 0.41
SST = SSTR + SSE
= 14.308 + 0.41
= 14.718
Step 3: Calculate the degrees of freedom:
– Total degrees of freedom (dftotal): n – 1, where n is the total number of observations.
– Treatment degrees of freedom (dfTreatment): k – 1, where k is the number of groups.
– Error degrees of freedom (dfError): dftotal – dfTreatment.
In our case, n = 30, k = 3, so:
Dftotal = 30 – 1 = 29
dfTreatment = 3 – 1 = 2
dfError = 29 – 2 = 27
Step 4: Calculate the mean squares:
– Treatment mean square (MSTR): SSTR / dfTreatment
– Error mean square (MSE): SSE / dfError
MSTR = 14.308 / 2 = 7.154
MSE = 0.41 / 27 ≈ 0.015
Step 5: Calculate the F-statistic:
The F-statistic is the ratio of the treatment mean square to the error mean square.
F = MSTR / MSE = 7.154 / 0.015 = 476.933
Step 6: Determine the critical value and compare with the F-statistic:
Using the F-distribution table or statistical software, we can find the critical value for a given significance level and degrees of freedom. Let’s assume a significance level of α = 0.05.
Looking up the critical value for α = 0.05, dfTreatment = 2, and dfError = 27, we find the critical value to be approximately 3.354.
Since the calculated F-statistic (476.933) is much larger than the critical value (3.354), we can reject the null hypothesis.
Step 7: Interpretation:
Based on the results, we can conclude that there is a significant difference between the means of the scores of early, middle, and late adolescents on the adjustment scale.
In summary, the ANOVA analysis indicates a significant difference among the groups, suggesting that the scores on the adjustment scale vary significantly across the different stages of adolescence.
