# Which group of dieters in this study had the lowest all-cause mortality rate? can this be determined without a t-test?

I'm having trouble determining whether between the vegan and pescatarian group which had the lowest all-cause mortality rate given the results from the study and whether this inference can be made without a t-test.

Results There were 2570 deaths among 73 308 participants during a mean follow-up time of 5.79 years. The mortality rate was 6.05 (95% CI, 5.82–6.29) deaths per 1000 person-years. The adjusted hazard ratio (HR) for all-cause mortality in all vegetarians combined vs non-vegetarians was 0.88 (95% CI, 0.80–0.97). The adjusted HR for all-cause mortality in vegans was 0.85 (95% CI, 0.73–1.01); in lacto-ovo–vegetarians, 0.91 (95% CI, 0.82–1.00); in pesco-vegetarians, 0.81 (95% CI, 0.69–0.94); and in semi-vegetarians, 0.92 (95% CI, 0.75–1.13) compared with nonvegetarians.

• It is not clear why you would want to avoid a formal test (t or otherwise). However, the CIs for the two groups you mention overlap substantially. I have no idea how the CIs were computed, but if they are valid, it seems that a formal test would not find a significant difference between these groups. Jul 31, 2019 at 0:04
• Yes there is an overlap but as i understand it the null can still be rejected by constructing one distribution for the difference in mean between groups. If the 95% CI doesn’t contain 0, then there is a statistically significant difference between groups. I did the t-test and i think the null hypothesis can be rejected. Data for Group A: 0.65 (0.43–0.97) Data for Group B: 0.90 (0.60–1.33)
– Matt
Jul 31, 2019 at 22:36
• You may be mixing criteria for one vs. two sample tests. CI(.43,.97)$doesn't contain 0, so mean .65 differs significantly from 0. This CI overlaps the CI (.60, 1.33). In fact, each CI contains the center of the other. Roughly, I'd call that 'substantial' overlap. So, without seeing your data, I'm still guessing there is no signif difference btw groups. Original question is missing now, so I'll leave it to you to sort this out. But do see example in my 'Answer'. Jul 31, 2019 at 23:08 • Please do not vandalize your question. When you posted on SE, you gave up ownership of the content under CC BY-SA 3.0. If there are no answers, you may delete your own question (see here ): just click the faint gray 'delete' at lower left (your account needs to be registered for this). Otherwise, the thread will remain according to SE's rules. Aug 1, 2019 at 0:58 ## 1 Answer Comment continued: Generate fake data: Two normal samples, each of size 30. Respective 95% CIs $$(47.4,\,53.4)$$ and $$(51.4,\,58.1)$$. Overlapping, but not as substantially as yours. (Centers of CIs, 50.4 and 54.7, are not contained in the CI for the other sample.) set.seed(731) x1 = rnorm(30, 50, 7); x2 = rnorm(30, 55, 7) t.test(x1)$$conf.int; t.test(x2)$$conf.int [1] 47.42130 53.41585 attr(,"conf.level") [1] 0.95 t.test(x2)$conf.int
[1] 51.37631 58.10891
attr(,"conf.level")
[1] 0.95


Welch two-sample t test does not (quite) show significance at the 5% level.

t.test(x1,x2)

Welch Two Sample t-test

data:  x1 and x2
t = -1.9621, df = 57.235, p-value = 0.05462
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-8.73668401  0.08861816
sample estimates:
mean of x mean of y
50.41858  54.74261