Evaluate the expression when y=1 expression: 5y
1 answer:
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Answer:
Step 1: The null and alternative hypothesis for the one way analysis of variance is,
H₀ : There is no difference in the means of each area.
H₁ : At least one mean is different.
The objective of this question is to test whether the income of the three areas are significantly different.
Step 2: The problem states to use 0.05 significance level.
Step 3: The test statistic follows the F-distribution.
Step 4: Decision rule is based on the critical value.
Numerator degrees of freedom is,
K - 1 = 3 - 1
= 2
Denominator degrees of freedom is 2.
n - k = 12 - 3
= 9
Using the F-distribution table for ∝=0.05 with numerator and denominator degrees freedom, the critical value is 4.26. The decision rule is to reject H₀ if the computed value of F exceeds 4.26.
Step 5: Calculations.
Using Excel follow the steps mentioned below.
1. Import or type the data into spreadsheet.
2. Data--------> Data Analysis----------> Anova: Single Factor, click OK.
3. Select Input Range, make a tick mark on Labels in the first row, enter 0.05 for alfa.
4. Click OK.
The obtained output for the one-way ANOVA is,
[ find the figure in attachment]
From the obtained output, the computed test statistic value, F = 14.18 , which is greater than the critical value of 4.26. So, reject H₀
Step 6: Interpretation. The average income of each area is different. It can be concluded that the average income is significantly different .
