I am trying to check if my sample characteristics are similar to the population characteristics for the following 2 cases.
1. I have the information regarding the gender distribution of my population of employees in a company (male:1577 and female:500). I want to confirm that this distribution is not statistically different from my sample's gender distribution (male:124 and female:51). What statistical test should be appropriate in this regard?
2. I have one more information about my population- the percentage of employees under 3 different categories of tenure.
(0-5 years=40%; 5-10 years=34%; above 10 years=25%).
For my sample I have the tenure information in 4 categories
(0-2years=17.7%; 2-5 years 28%; 5-10 years= 33.7%; above 10 years= 20.6%)
In this case will it be appropriate if I add the percentage in 0-2 years (17.7%) and 2-5 years(28%) and consider it as 0-5 years group (17.7%+28%= 45.7%) and then do a statistical analysis to check if the sample tenure distribution is different from that of the population?
If yes then what kind of statistical test should be used to check if there is a difference between the sample and population in terms of tenure in this case?
I am sorry if my explanation is confusing but it would be very kind of you if I can have some idea.
? You already know the gender distribution of your population:
proportion of male:1577/2077 and proportion of female:500 /2077.
you have taken a sample from this population your self?
Then why or what you want to check?
For both cases z-test for proportions would be suitable. You will be testing differences between proportions (relative frequencies) of categories (male/female, each of tenure categories). You can check different modes of sample proportion inferences here: https://homepage.divms.uiowa.edu/~mbognar/applets/p.html
David Morse Thank you so much for your kind explanation and the worksheet! Its been really a great help! I will study the worksheet for better understanding. Thank you so much again!
Hi Roksana Rezwan and David Morse .
Must be something wrong with the calculation here, try f.ex. Quick Calcs and check whether your sample is similar to your overall composition of your company staff
https://www.graphpad.com/quickcalcs/contingency1/
See attached.
I.e., the significance test does not support that the groups are dissimilar.
Best
John-Kåre
John-Kåre Vederhus Thank you so much for your kind help! I learned a lot from your shared material!
Roksana Rezwan
I suppose the reason for doing this analysis is that you have a convenience sample (those who answered the survey) and wanted to see whether they are "representative" for the company at large?
Best
JK
Hello all,
The calculation that John-Kåre Vederhus is not a goodness of fit test, it is a test of independence (a different hypothesis altogether). That is why it does not match with the ones I furnished.
Cheers,
John-Kåre Vederhus Yes that is true. My sample is a convenience sample.
Hi Roksana Rezwan . David argued well for his solution and his excel sheet was fine. David Morse , is the goodness of fit test included in standard statistical packages, like for example SPSS? Seems like it should be simple to calculate it from a 2 x 2 crosstab....
Best
JK
Hello John-Kåre Vederhus and Roksana Rezwan ,
The problem with the 2 x 2 contingency table set-up (traditional chi-square test of independence) is that the four cells are not mutually exclusive. All males who appear under the "Sample respondent" category also are a part of the "Population/Company" category. Likewise for females.
SPSS does include a goodness of fit test, but you have to set it up specifically for a given situation (unless you're happy testing the null hypothesis that the category frequencies should be equal). Analyze / Nonparametric tests / One-sample, then choose "Custom" and go from there. (Or, Analyze / Nonparametric tests / Legacy dialogs... then choose Chi-square.)
Now, for a mea culpa; D'oh! I forgot to divide the squared cell deviations by expected cell frequencies, so the results are different:
For sex of sample (vs. population distribution): Chi-square(1, N = 175) = 2.461, p = .117. No difference in sample distribution by sex from population/company.
For years of job tenure (vs. population distribution): Chi-square(2, N = 175) = 2.806, p = .246. No difference in job tenure distribution from population/company.
Worksheet (new, improved, more likely correct) is attached.
Cheers,
David Morse Thank you so much for the kind update and all the calculations! Its been really helpful a lot!
Best wishes
Roksana
Subrata Chakraborty Thank you for your reply. Yes I collected the sample from the population by myself, but I could not do randomization during my survey distribution. So it became more of a convenience sample. So I just wanted to check (if it is necessary) ,if my sample represents the population by checking if the characteristics of the sample are similar (statistically significantly) to the population.
Roksana Rezwan . Technically, they are not "similar", they're just not "dissimilar" enough according to the fit analyses David Morse suggested (the statistic doesn't support that they are dissimilar), and you can count them as representative for the employees on the variables you checked for. There are probably still some bias that you should mentioned in your limitation section. E.g. if your study is about "satisfaction" or other sensitive information, those who are dissatisfied will probably be less prone to answer the survey (for example; they may be afraid to be recognized). Not everything can be solved with statistics :-)
Good luck with your work.
Best
John-Kåre
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