Statistics – Chi Square Test of Independence and Confidence Intervals Explored

chi-squared-testconfidence interval

Can you calculate confidence intervals when performing a chi square test of independence. Someone asked me to provide CI but I'm not sure how this is possible since the test only tests for indepedence.

Im comparing countries and whether or not they are single offenders or repeatable offenders. I found that a 5% level of siginficance to reject the null hypothesis. What is there to say more than there is an association between both variables.

I mean I could also measure that association with phi or cramers v but that's all I can really provide?

Am I wrong ?

Best Answer

The correct response would be, "Confidence intervals for what statistic [estimate of what parameter] ?"

Probably the most useful would be providing the confidence interval for phi (or Cramer's V), or for a meaningful odds ratio for the table.

Beyond that, you can provide whatever information is useful for the readers. Maybe, the proportions within each row or column ? Or the ratios of these ?

Related Solutions

Solved – Chi Square test for survey data

It appears that you are first doing an omnibus test (Chi square test for independence) with 2 df to determine if the "like status" and "gender" are independent or not. And then you are doing post-hoc tests on the individual rows (Chi square goodness of fit tests) to see if the males/females are equally likely under each row. According to This Link under the section "Post Hoc Follow-up Tests", these post-hoc tests are allowable. Each row would generate a Chi square test with 1 df. They would test, for instance "Ho: men and women 'are likers' at the same rate", for each row.

However, I am leery that no adjustment was made for multiple comparisons. Since it appears you are doing three of these 1 df tests, you should adjust your $\alpha$ to correct the familywise error rate (Bonferroni correction for instance).

If your client wants to know how much more likely men are to be a "liker", etc. you could (a), provide a point estimate based on your data as Peter Flom suggested, or (b) you could construct a CI for the difference between the two proportions if you want an interval estimate. Along with the statement that the difference is significant (or not significant), my guess is that a point estimate would suffice for your clients.

Other than the problem with not controlling the familywise error rate, the analysis seems adequate to me. I hope this helps.

Solved – Chi-square test statistic is significant but $p$-value says otherwise – what’s the correct interpretation

If your test statistic follows $\chi^2_1$ distribution, then you have to look at the probability of observing value $0.065$ or greater value under this distribution. Using R software to get the values you can find that probability of observing such or greater value is $0.79$ under null distribution:

> 1-pchisq(0.065, df = 1)
[1] 0.798761

so obviously $0.79 \not \le 0.05$. On the plot below you can see your value marked by blue lines and the critical value that is marked by red lines.

The critical value is

> qchisq(.95, df = 1, lower.tail = TRUE)
[1] 3.841459

under $\chi^2_1$ distribution $5\%$ cases are greater and $95\%$ are lower than this value. You made mistake of looking at value

> qchisq(.05, df = 1)
[1] 0.00393214

i.e. a point where $5\%$ are less than this value and $95\%$ are more than this value, so you were comparing to $p=0.95$ rather than $p=0.05$.

Check also What is the meaning of p values and t values in statistical tests? to learn more about how to understand $p$-values.

Best Answer

Related Solutions

Solved – Chi Square test for survey data

Solved – Chi-square test statistic is significant but $p$-value says otherwise – what’s the correct interpretation

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