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Statistics is not a normal part of my job. I've been given survey data from three years. Survey data includes a number of categories the respondents are split into. There is a set of five possible responses for each. Each year, the percentage of each response for each category is calculated. I need to provide a three year average for each response for each category.

Which method do I use to calculate the three year average for each category?

  1. Add the raw totals from each year for each response and divide by the total respondents over the three years.
  2. Add the three single-year percents, then divide by 3 to get an average percent.

I can provide example data if that would be helpful.

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You can do either, but they mean different things. Your method 2 give each year the same weight, and an individual's weight depends on the size of the sample that year. The first method gives each individual the same weight. Most textbooks would give the first method, because it is "optimal" if nothing has really changed from one year to the next. I would probably do the second, since I don't trust the passage of time to leave things unchanged.

If the samples were reasonably large in all three years, it won't matter much either way.

How were the original means calculated? Were the surveys complex? Was there stratification?

You didn't say anything about a confidence interval, but the calculation of a confidence interval will depend on which method you choose.

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  • $\begingroup$ Each year the survey remains the same, but covers a new set of people. The categories range from 0-50 people in each, example categories might be different age ranges. We are only assessing one question, that has five possible responses. The single-year percents are just #of response x / total people in category. Does that help or change the recommendation? $\endgroup$
    – ValerieN
    Dec 18, 2015 at 16:19
  • $\begingroup$ since the numbers are comparable each year and the samples are small, I would go with method 1. it is slightly more efficient. $\endgroup$
    – Placidia
    Dec 18, 2015 at 16:31

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