A real estate developer is considering investing in a shopping mall on the outskirts of Atlanta, Georgia. Three parcels of land are being evaluated. Of particular importance is the income in the area surrounding the proposed mall. A random sample of four families is selected near each proposed mall. Following are the sample results. At the 0.05 significance level, can the developer conclude there is a difference in the mean income

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 .

answer: procent 54 wynosi 33,33%

if x = 1, then the answer is 3.5