In order to determine whether we should reject or not reject the claim that there is no difference in population mean annual profits per employee in each of the four types of companies (computers, aerospace, heavy equipment, and broadcasting), we can conduct a one-way ANOVA (Analysis of Variance) test. The first step of this test is to calculate the Total Sum of Squares (SSTOT).

Here’s the step-by-step process for calculating SSTOT:

Step 1: Organize the Data

You provided data for four categories: computers, aerospace, heavy equipment, and broadcasting. We will use the following values:

• Computers: 27.4, 23.3, 14.3, 8.2, 11.7
• Aerospace: 13.8, 9.7, 11.9, 8.8, 6.2, 19.4
• Heavy Equipment: 22.2, 20.3, 7.7, 12.6, 7.5
• Broadcasting: 17.2, 16.8, 14.1, 15.8, 10.4, 9.5

Step 2: Calculate the Grand Mean ($\overline{X}$)

The grand mean is the mean of all the data points combined, regardless of the group they belong to.

$\overline{X} = \frac{\text{sum of all data points}}{\text{total number of data points}}$

Step 3: Calculate SSTOT

The formula for the Total Sum of Squares (SSTOT) is:

$SSTOT = \sum (X_{ij} - \overline{X})^2$

Where:

• $X_{ij}$ represents each individual data point.
• $\overline{X}$ is the grand mean.

Now, let’s calculate the grand mean and SSTOT.The Total Sum of Squares (SSTOT) is approximately 690.391 (rounded to three decimal places).

Would you like to continue with the rest of the ANOVA test or ask any questions?

Here are 5 related questions to expand on this:

1. How do you interpret the result of SSTOT in the context of ANOVA?
2. What is the next step after calculating SSTOT in the ANOVA process?
3. How is the between-group variation (SSB) calculated in ANOVA?
4. Can you explain the importance of the F-statistic in ANOVA?
5. How does the p-value in ANOVA relate to the level of significance?

Tip: In ANOVA, the higher the SSTOT, the more variation exists in the data. However, it's the comparison between the within-group and between-group variations that helps determine statistical significance.