My Math Forum Probability re Placebo

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 September 2nd, 2017, 01:43 PM #1 Newbie   Joined: Sep 2017 From: Oxford Posts: 1 Thanks: 0 Probability re Placebo Any help appreciated to calculate the below. A clinical trial has two arms. Arm 1 is placebo. The results generated were 17 responders out of 44 patients (38.6%) Arm 2 is the active trial. The results generated were 49 responders out of 90 patients (54.4%) How do I calculate the p number to show the likelihood of the active arm obtaining that statistical result by chance, i.e. the chance of there being no real difference between the active arm and placebo? Anything under p=0.05 will be statistically significant, which is what I am trying to assess. Thanks in advance Last edited by siciliankan; September 2nd, 2017 at 01:44 PM. Reason: Extra information
 September 10th, 2017, 05:17 AM #2 Member   Joined: May 2013 Posts: 33 Thanks: 1 so lets create the probability square. Code:  prob(react to placebo) | prob(not react to placebo) | prob(react to actual) | |49/90 --------------------------------------------------------------------------------------- prob(not react to actual) | |41/90 ---------------------------------------------------------------------------------------- 17/44 | 27/44 | 1 we're looking for the upper left, the probability of reacting to placebo, and actual, which is simply 17/44*49/90 = 833/3960 =0.21 approx.
October 27th, 2017, 12:25 PM   #3
Newbie

Joined: Oct 2017
From: US

Posts: 11
Thanks: 0

Just put the following numbers into this chisquare calculator

17 27
49 41

you will get a p value of 0.085631065798, which is larger than 0.05.
Therefore, it is not significant at 0.05 level and you can not reject the null hypothesis of independence that placebo and trial are the same in 'react or not'.
Attached Images
 iCalcu-com-chi-square-test-example4-min.jpg (18.3 KB, 3 views)

Last edited by James White; October 27th, 2017 at 12:43 PM.

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