Timeline for Which statistical test should I use with proportions/percentages?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 3, 2018 at 16:56 | comment | added | dbwilson | If you focus on the individual purchase, then you can easily test this using a mixed-effects logistic regression model. The DV would be whether the purchase is expensive, the IV whether they had an iPhone, and a random intercept for person (person treated as a cluster). If the variance for the person random effect is zero, then your assumption of independence is plausible. | |
| Apr 3, 2018 at 16:54 | comment | added | dbwilson | If you are willing to make that assumption that they are independent, then you can do the simple 2 by 2 chi-square suggested above. However, I don't find that assumption plausible. | |
| Apr 3, 2018 at 16:41 | comment | added | Ami | Thanks for the help. I'm just not sure if this is the same when I make the contingency table. Would it be possible to observe purchases and then for each purchase check whether user had an iPhone and whether the product was expensive? I might be wrong but it looks like they wouldn't be dependent in that case @dbwilson | |
| Apr 3, 2018 at 16:02 | comment | added | dbwilson | Each person has a different underlying true proportion for the ratio of expensive items they buy. Thus there is a person effect or intra-class correlation for person. Another way of thinking about this is that you have purchases nested (clustered) within persons. | |
| Apr 3, 2018 at 14:12 | comment | added | Ami | I can't use the t-test because my distribution is not normal. I have the data for each customer but I can also calculate the total number. I'm not sure what you mean by "the purchases made by a single person are not independent" @dbwilson I would think they are. I also checked the Chi-square test in IBM SPSS and it seems like it is possible to give the single samples which are later used to build a contingency table and calculate the test values | |
| Apr 3, 2018 at 13:38 | comment | added | dbwilson | But the purchases made by a single person are not independent, so counting them up with all other other purchases to simply get the overall total expensive and inexpensive purchases violates the independence assumption, does it not? | |
| Apr 3, 2018 at 13:28 | comment | added | Patrick Malone | @dbwilson I read the question as saying Ami had the total number of purchases by person, as well, which would make chi-square or logistic fine. If that's not correct, and Ami only has proportions at the individual level as raw data, then you're right. | |
| Apr 3, 2018 at 11:51 | comment | added | dbwilson | This treats to purchase as the unit-of-analysis and not the individual customer. If I understand the data correctly, Ami has a proportion for each customer, so there is a distribution of proportions within each group. Assuming your proportions are not close to 0 or 1 and your sample sizes are of a decent size, you can do a t-test. Alternatively, you could do a beta regression with customer as the IV. | |
| Apr 3, 2018 at 10:29 | comment | added | Ami | Thanks a lot for the detailed explanation. That was very helpful. | |
| Apr 3, 2018 at 10:28 | vote | accept | Ami | ||
| Apr 3, 2018 at 9:38 | history | answered | Patrick Malone | CC BY-SA 3.0 |