I need help with weighted averages

Shifty707 · 2014-01-09 06:34:59

Hi all.

I'm wondering if someone could help. I'm not very good with weighted averages nor could I figure out what weight I'm using. Can someone help me understand weighted averages more and help me solve weighted averages for the two images below?

Very much appreciated. Thank you.

Last edited by Shifty707 (2014-01-09 06:37:20)

SteveB · 2014-01-09 08:25:31

I think you have to take each pair of Score and Sample size and multiply them, and then add these together to get a total,
then divide by the total sample size for all of the rows you have added:

eg. (3.83 x 521) + ..... + ...... I have just given the first row of the scores x sample sizes (I won't add all of them because there are a lot - and also that is your job)....
Then add 521 + 806 + ..... and get the sample size total

Remember to treat each year separately of course for both totals in each case.
(Similar principle for the top table except that they are not years, and similar I think for the percentages,
just treat them like scores - multiply them by the sample sizes in each row - then sum them - then divide by the total sample size.
I notice that the total sample sizes are there already and that the averages are already given.)

Last edited by SteveB (2014-01-09 08:34:14)

Shifty707 · 2014-01-09 10:37:51

SteveB wrote:

I think you have to take each pair of Score and Sample size and multiply them, and then add these together to get a total,
then divide by the total sample size for all of the rows you have added:
eg. (3.83 x 521) + ..... + ...... I have just given the first row of the scores x sample sizes (I won't add all of them because there are a lot - and also that is your job)....
Then add 521 + 806 + ..... and get the sample size total
Remember to treat each year separately of course for both totals in each case.
(Similar principle for the top table except that they are not years, and similar I think for the percentages,
just treat them like scores - multiply them by the sample sizes in each row - then sum them - then divide by the total sample size.
I notice that the total sample sizes are there already and that the averages are already given.)

SteveB, thank you so much for your reply. It does help make sense of things. The only issue I want to know is how is everything weighted. I've researched online and seen examples with weights already inserted. That's the only thing I'm confused on. How do I figure the weight of each hotel?

SteveB · 2014-01-09 22:35:22

That is a good point. I had assumed that the sample figures were intended as the weights - in that they represented the 'units' of concern for the
purposes of the average. It does look like you don't have any data to 'weigh' the hotel so to speak. It depends on the context of what you are trying
to do. For instance if this is a very big hotel chain, then you might look at the turnover figures for the hotels, or the number of customers per year,
in general something that gives an idea of the scale of importance in relative terms for different hotels.
Why were the sample sizes different I wonder? I was assuming that the sample size varied in proportion to how 'big' the hotel was.
Perhaps it just depends upon how much data you have for each hotel and not necessarily the prominence of the hotel per se.
It looks like it is not strictly possible to obtain a weighted average with the information you have given, but it is common practice in statistics to
work out a weighted average by the multiplication that I described:
weighted average = sum of ((score of item) x (frequency of item)) / sum of (frequency of item)

Imagine that you had not summarised them in a table, but had all of the individual scores in a list then the above formula would give the same averages.

If the sample size is not the population size, and is not proportional to it, then it is not strictly speaking a frequency suitable for weighted averages.
On the other hand the scores were presumably worked out by dividing the total row score by each sample size for each row so it is
reasonable to get an overal average by reversing this division.

Last edited by SteveB (2014-01-10 02:44:47)

Math Is Fun Forum

#1 2014-01-09 06:34:59

I need help with weighted averages

#2 2014-01-09 08:25:31

Re: I need help with weighted averages

#3 2014-01-09 10:37:51

Re: I need help with weighted averages

#4 2014-01-09 22:35:22

Re: I need help with weighted averages

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