**Basic Concepts**

The one random factor structural model is given by the formula

As usual *ε _{ij}* can be consider to be a random variables with variance , but unlike in the fixed factor case, this time

*τ*can be considered to be a random variable (instead of a constant) with variance .

_{j}We assume that for all *i* and* j, * and are pairwise independent and

Since and are independent it follows that

We test the null and alternative hypotheses

H_{0}: = 0 H_{1}; ≥ 0

Note that the null hypothesis is equivalent to ≤ 0. If the null hypothesis is true, then variability among random selected treatments is simply due to random error (i.e. ). If the alternative hypothesis is true then variability among random selected treatments is significantly higher than just the random error.

The definitions of *SS _{T}, SS_{B}, SS_{E},* and similarly for the

*MS*and

*df*terms, are exactly the same as for the one fixed factor model described in One-way Anova Basic Concepts. In fact the calculations are also the same, although the interpretation of the results is different.

Using the notation for one-way ANOVA with fixed effects

For a balanced model *m = n*. If the null hypothesis is true, then = 0, and so

If the alternative hypothesis is true, then ≠ 0, and so

**Estimates of Variances**

We first note that can be estimated by *MS _{E}* and can be estimated by

(provided *MS _{B} > MS_{E}*). With a balanced model, we can construct a confidence interval for based on

Assuming a balanced model, we can construct (Satterthwaite) a confidence interval for based on

As usual the population mean can be estimated by the grand mean, *μ* = *E*[*x̄*]. The variance of this estimate is (*m* + )*km*, which can be estimated by *MS _{A}*/

*km*.

**Data Analysis**

**Example 1**: A biologist is trying to determine whether toxicity of a certain of mushroom is the same across the 100 different regions being studied. She takes a random sample of 5 of these regions and then measures the toxicity of 10 mushrooms selected at random in each of the selected regions. Determine whether there is a significant difference in the toxicity of these mushrooms in different regions based on the data on the left side of Figure 1.

**Figure 1 – One Random Factor Anova dialog box**

To run the analysis you can use Excel’s **Anova: Single Factor** data analysis tool, as described in One-way Anova Basic Concepts. Alternatively you can use the Real Statistics **One Factor Factor ****Anova **data analysis tool, as follows, since it provides some additional information.

For Example 1, enter **Ctrl-m** and double click on **Analysis of Variance**. Next select **Anova: one factor **from the dialog box that appears.** **Fill in the dialog box that now appears as shown on the right side of Figure 1. This is the same procedure that is used for one fixed factor Anova, except that you can check the **Random Factor** option.

The output is shown in Figure 2.

**Figure 2 – One Random Factor ANOVA data analysis**

The first part of the analysis is identical to the output when the **Random Factor** option is not checked. As we can see from Figure 2 cell L13, p-value = .0165. Assuming that α = .05, we conclude there is a significant difference in the toxicity of the mushroom under study among the 100 regions.

The second part of the output provides an estimate of the mean toxicity of the mushroom across the 100 regions (cell H23) and a 95% confidence interval (cells K23 and L23), as well as estimates of the population between groups variance and error variance (cells H19 and H20) and their 95% confidence intervals (cells L19, L20 and M19, M20).

**Caution**: As mentioned previously, the above estimates assume that *MS _{B} > MS_{E}*. When this is not the case, a negative variance will occur, which is of course impossible, and so the estimates should be ignored. Also if the degrees of freedom for the between groups variance (cell I19) is low then the confidence interval will be large, and so will not be very useful. In fact if this

*df*is less than one, Excel will round it down to zero when calculating the chi-square values (cells J19 and K19) resulting in an error.

Charles,

In your example above, your software calculates cell I23 by H19/(nk)– it should have been MSB/(nk), or cell J13/(nk) = 12.44.

Heather,

You are correct. I have just corrected the referenced webpage. I will correct the examples workbook when I release a new version in a few days. Thanks for catching this mistake.

Charles