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Is it chance? Use a T-Test to identify how likely an intervention worked (edit)
Revision as of 14:00, 4 March 2022
, 14:00, 4 March 2022→Practical examples
The p-value in this example, <code>0.24</code>, is above <code>0.05</code> and therefore we will assume that any difference in sleep between the nights is due to pure chance.
The p-value in this example, <code>0.24</code>, is above <code>0.05</code> and therefore we will assume that any difference in sleep between the nights is due to pure chance.
=== t-tests in R ===
You can do the same t-test in the statistical programming language [[R]], using the same example:
<pre>
melatonin <- c(8.53,7.64,7.26,7.53,7.85) # load intervention data
no_melatonin <- c(7.91,7.7,7.7,7.13,7.62,7.51) # load control data
t.test(melatonin,no_melatonin,alternative='greater',paired=FALSE) # run the t-test
</pre>
First we load both data series into R and then run the actual t-test. With <code>alternative='greater'</code> we specify a one-sided t-test in which we test whether the values of the intervention data are larger than those of the control. Alternatively we could have chosen <code>alternative='less'</code> or <code>alternative='two.sided'</code>. With <code>paired=FALSE</code> we specify that these are independent, non-paired samples.
Running our test results in the following output:
<pre>
Welch Two Sample t-test
data: melatonin and no_melatonin
t = 0.69689, df = 5.9576, p-value = 0.2561
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
-0.2992509 Inf
sample estimates:
mean of x mean of y
7.762 7.595
</pre>
We see that <code>p-value = 0.2561</code>, indicating that the sleep-duration when taking melatonin is not substantially larger compared to the control and that any differences are likely just by chance.
== Limitations ==
== Limitations ==