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July 7

Is there a better way to measure budget items in surveys than the median?

If you use the mean average in surveying people about budget items, outliers can have a disproportionate influence, but multiple choice has low resolution and allows leading questions. Therefore the median is preferable. Is the same true for surveying people's opinions about changes in budget items? 99.24.223.58 (talk) 00:06, 7 July 2011 (UTC)[reply]

Could you give an example of the sort of question you have in mind? People in general are so ignorant about numbers that I doubt even median values are useful. Looie496 (talk) 00:31, 7 July 2011 (UTC)[reply]

I went to a "town hall" style meeting where people wanted to cut a certain budget item 20%, but only 5% and 10% were available choices. When asking people how much money they would budget for an expense, and how much they would change the budget for an existing expense, is there any better method than using the median for the result? 99.24.223.58 (talk) 03:00, 7 July 2011 (UTC)[reply]

I'm not sure this is the right answer to the actual question you have in mind, but there are various sorts of trimmed means that are compromises of a sort between the mean and the median. Michael Hardy (talk) 05:16, 7 July 2011 (UTC)[reply]

I can't really see the point in getting an average budget saving, it's the amount of money saved that matters so what's the point of saving 10% on lots of tiny things and getting the mean or mode or median of that with one big one of 5%? By the way I approve of the restriction to 5% and 10%. As a general rule of thumb a budget increase or decrease of 15% or more needs to be specially managed, preferably by replicating or removing entire units. Changing a single unit by that much causes difficulty and inefficiency. Dmcq (talk) 18:19, 7 July 2011 (UTC)[reply]

With multiple choice, who gets to pick the choices? There is an important sampling theorem which states that response options should bracket expected responses for validity, so is that rule of thumb documented anywhere? 208.54.5.188 (talk) 00:50, 8 July 2011 (UTC)[reply]

Does anyone know whether the outcomes of numerical surveys by trimmed means are in general more or less accurate than by median? I imagine market makers in securities and sports gambling might be interested. 99.24.223.58 (talk) 21:49, 8 July 2011 (UTC)[reply]

If it helps, I read an article in Dr Dobbs Journal years ago (article was still available on the internet recently) saying that the median is better for removing the noise from a noisy signal, than the mean is. 2.101.12.198 (talk) 11:07, 10 July 2011 (UTC)[reply]

Note a related issue. By disallowing 20%, the survey assumes well-ordered preferences. That is, if a person's top choice is 20%, then his second best choice is 10%. That may not be the case. For example, I may prefer that a certain program either be cut (20%) or supported (5%), but not half-heartedly done (10%). Thus, 20% could be the top choice for everyone in the room but, by forcing a choice between 5% and 10%, you could get half the people preferring 5% and half preferring 10%. Wikiant (talk) 14:45, 10 July 2011 (UTC)[reply]