0
$\begingroup$
> prop.test(table(D2$SMOKE,D2$RACE),correct=FALSE)

2-sample test for equality of proportions without continuity correction

data:  table(D2$SMOKE, D2$RACE)
X-squared = 23.26, df = 1, p-value = 1.415e-06
alternative hypothesis: two.sided

95 percent confidence interval:
 0.1863890 0.3809326

sample estimates:
   prop 1: 0.7733333   
   prop 2:  0.4896725 
$\endgroup$
  • $\begingroup$ what is your hypothesis? How do you compute proportions ? IS there any logic to use chi- square statistic for your study ? $\endgroup$ – Subhash C. Davar Nov 15 '17 at 16:05
0
$\begingroup$

prop1 and prop2 are the probabilities of success for your 2 groups. the p-value is less than one minus the level (95% or 0.95) of the confidence interval, which indicates the proportions of the characteristic studied are statistically significantly different in the 2 groups.

maybe you can have a look at this tutorial which show a complete example of such test.

$\endgroup$
  • 1
    $\begingroup$ Generally, we also ignore the $\chi^2$ (X-squared) value and the df (degrees of freedom) since the confidence interval contains all the information we need. We can also conclude that the proportion difference is statistically significantly different because the confidence interval does not contain 0 (indicating no proportion difference). $\endgroup$ – AdamO Nov 14 '17 at 19:00
  • $\begingroup$ It is incorrect to use chi-square test for the stated purpose . Moreover, question does not indicate sample -sizes and requisite classification tables or data that could be useful for suggesting an apprpriate test - t or z etc. $\endgroup$ – Subhash C. Davar Nov 15 '17 at 16:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.