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# One-way analysis of variance

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How do I figure this problem out using the population means?

The manager of a computer software company wishes to study the number of hours top executives spend at their computer terminals by type of industry. The manager selected a sample of five executives from each of three industries. At the .05 significance level, can she conclude there is a difference in the mean number of hours spent at a terminal per week by industry?

Banking----Retail-----Insurance
12------------8------------10
10------------8------------8
10------------6------------6
12------------8------------8
10------------10-----------10

https://brainmass.com/statistics/analysis-of-variance/one-way-analysis-variance-31627

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This solution shows how to conduct a one-way analysis of variance with all calculations done by hand. All formulas, calculations and complete explanations are included. Please see the attached Word document for the solution to this exercise.

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How do I figure this problem out using the population means?

The manager of a computer software company wishes to study the number of hours top executives spend at their computer terminals by type of industry. The manager selected a sample of five executives from each of three industries. At the .05 significance level, can she conclude there is a difference in the mean number of hours spent at a terminal per week by industry?

Banking----Retail-----Insurance
12------------8 ------------10
10------------8 ------------8
10------------6 ------------6
12------------8 ------------8
10------------10 ----------- 10

Solution

This is an example of an analysis of variance problem with a completely randomized design. A completely randomize design is a design for which independent random samples of experimental units are selected for each treatment.

Let be the population mean number of hours top Banking executives spend at their terminals, be the mean hours for Retail executives, and be the mean for Insurance executives. The goal is to use the sample data to make an inference about these population means. We want to test the hypothesis that all three means are the same vs. the alternative that at least one differs from the others.

At least one mean differs from the others

First calculate the total sum of squares for the data. This is done by calculating the sample variance of all 15 measurements in the data set and multiplying by n - 1 = 14. Or equivalently simply calculate the numerator of the sample variance. Recall that the formula for the sample variance is:

The total sum of squares is the numerator of the sample variance:

Since there are n = 15 measurements in this calculation, there are n - 1 = 14 total degrees of freedom in this data set.

Next, calculate the Treatment or "Factor" sum of squares. The formula for this is:

Where:
The sample size of each factor level (5 in this case for all 3 groups)
The sample mean of the ith group.
The overall mean of the 15 members of the data set = 9.07

In this problem

So now it's easy to get the Treatment (or "Factor") sum of squares:

There are p=3 treatments (aka "factor levels") here so there are p - 1 = 2 degrees of freedom for treatment.

Next create an ANOVA table using the values we have already calculated..

Analysis of Variance
Source DF SS MS F P
Factor 2 22.93 11.47 5.73 0.018
Error 12 24.00 2.00
Total 14 46.93

The values we calculated above are entered in bold type.

The degrees of freedom and sums of squares for in the "Error row" were obtained by subtracting the value in the "Factor" row from the "Total" row.

The values in the MS (or mean square) column were obtained by dividing the sums of squares by degrees of freedom in the same row.

The F value is the Mean Square for "Factor" divided by the Mean Square for "Error".

This F value is the test statistic. If the null hypothesis is true, this test statistic has an F distribution with 2 numerator degrees of freedom and 12 denominator degrees of freedom.

The p-value for this test is 0.018 (the rightmost element in the ANOVA table), the probability that a randomly selected member of the F distribution with 2 and 12 degrees of freedom is greater than the calculated test statistic value of 5.73. This is the probability of observing a test statistic as extreme or more so than the one we calculated under the assumption that the null hypothesis is true. Since this value is less than the significance level of 0.05, there is sufficient evidence to reject the null hypothesis and conclude that the alternative is true. There is sufficient evidence to show that at least one of the three population means differs from the others.

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