Statistics/Maths Help

[GrandOldTeam]

I have two sets of data that I need to average.

First set consists of 200, the next 37.

Initially, I foolishly just added up the statistics and averaged, but obviously the first set skews the overall data.

So, I thought I would do the same but use the % of each data. To an extent, this does the same.

Its been years since I done this stuff and struggling!

Whats the best way? NEBB!!??

 

[Micky]

You could do the mean of both sets of data, but wouldn't know how to get the average for both sets together.

Did abit of this at college and the way your doing it is the way I did it. Got told that because it is misleading with the first bit of data, you couldn't use it.

*i'm not much help, i know*

 

[GrandOldTeam]

Basically, it would be like asking 200 fans in England;

Who do you support
1. Everton 152 (76%)
2. Arsenal 44 (22%)
3. Tottenham 4 (2%)

And asking 37 fans in a village in Africa;

Who do you support
1. Everton 35 (95%)
2. Arsenal 2 (5%)
3. Tottenham 0 (0%)

I need a way to average them both, without the majority of the first sample skewing data. There will be an easy solution, just cant think!

 

[Dom1878]

Standard Deviation of boht sets of Data.

 

[GrandOldTeam]

Standard Deviation of boht sets of Data.

Nah, wont touch SPSS or anything like that. Had to get the help of a young lady to help with Excel! Useless at stuff like that

 

[GrandOldTeam]

Nebb, Bruce or TX will sort this.

 

[Reidy's Bottle Of Grecian]

the sum = average(the average + the average)

 

[GrandOldTeam]

Say that again....

 

[Reidy's Bottle Of Grecian]

the sum = average(the average + the average)

add the averages together and divide by 2 as there are only 2 averages being added together.

or use logarithms.

 

[The Esk]

you need to weight the average.

So if the averages are A and B respectively, then the weighted average in this case will be:

(A x 200) + (Bx37)
------------------
237

 

[Reidy's Bottle Of Grecian]

ask carol

 

[GrandOldTeam]

you need to weight the average.

So if the averages are A and B respectively, then the weighted average in this case will be:

(A x 200) + (Bx37)
------------------
237

Cheers, I think I have done this. To be sure, any chance you could answer this?;

-------------
Prostitutes (37 sample)
5. If a prostitution tolerance zone was to be introduced, where should such an area be located?

Inside the City Centre 31 (84%)
Outside the City Centre 6 (16%)

Public (200 sample)
5. If a prostitution tolerance zone was to be introduced, where should such an area be located?

Inside the City Centre 115 (58%)
Outside the City Centre 85 (42%)
------------

From all 237, what is the % of;

Inside the City Centre
Outside the City Centre

 

[The Esk]

Cheers, I think I have done this. To be sure, any chance you could answer this?;

-------------
Prostitutes (37 sample)
5. If a prostitution tolerance zone was to be introduced, where should such an area be located?

Inside the City Centre 31 (84%)
Outside the City Centre 6 (16%)

Public (200 sample)
5. If a prostitution tolerance zone was to be introduced, where should such an area be located?

Inside the City Centre 115 (58%)
Outside the City Centre 85 (42%)
------------

From all 237, what is the % of;

Inside the City Centre
Outside the City Centre

OK:

Inside the City Centre = (31+115) / 237 = 61.6%
Outside the City Centre = (6+85)/237 = 38.4 %

 

[GrandOldTeam]

Ah great. Was doubting myself. Cheers esk!

 

[davek]

Is this for your degree? What are you taking, Applied Knockingshopology? :hay: