Question

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 .05 significance level, can the developer conclude there is a difference in the mean income? Use the usual six-step hypothesis testing procedure.

Answer 1

Step 1: The null and alternative hypothesis for the one way analysis of variance is,

There is no difference in the means of each area.

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,

Denominator degrees of freedom is 2.

Using the F-distribution table for with numerator and denominator degrees freedom, the critical value is 4.26. The decision rule is to rejectif 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. DataData 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,

Figure (1)

From the obtained output, the computed test statistic value, , which is greater than the critical value of 4.26. So, reject.

Step 6: Interpretation. The average income of each area is different. It can be concluded that the average income is significantly different .