## One-sample t-test [General Sta­tis­tics]

Hi Loky do,

❝ If I have a bioavailability study and want to compare its PK results with published data, is there any specific statistical test that could be used to prove there is no significant difference between both results?

If you have the subject’s data of the previous study, you could work with log-transformed values and use a paired t-test.
If not, i.e., you have only the mean ($$\small{\mu}$$), you could use a one-sample t-test, where $$\small{t=\frac{\overline{x}-\mu}{s/\sqrt{n}}}$$ with $$\small{df=n-1}$$.
Example in :

 set.seed(1234567) x <- rnorm(n = 24, mean = 20, sd = 5) # give a vector of your study data instead summary(x)    Min. 1st Qu.  Median    Mean 3rd Qu.    Max.   11.11   16.98   20.09   19.82   23.15   27.71 t.test(x, mu = 20, conf.level = 0.90)         One Sample t-test data:  x t = -0.1917, df = 23, p-value = 0.8497 alternative hypothesis: true mean is not equal to 20 90 percent confidence interval:  18.25799 21.39151 sample estimates: mean of x  19.82475 t.test(x, mu = 25, conf.level = 0.90)         One Sample t-test data:  x t = -5.6612, df = 23, p-value = 9.191e-06 alternative hypothesis: true mean is not equal to 25 90 percent confidence interval:  18.25799 21.39151 sample estimates: mean of x  19.82475

Problems:
• You have to assume that $$\small{\mu}$$ is the true mean without any error. Strong assumption, likely false.
• Based on normal distributed data. Likely the reported $$\small{\mu}$$ is the arithmetic mean. If you have also its standard deviation, you could try to bootstrap values from a lognormal distribution. Not trivial.

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Helmut Schütz

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