Suppose you are conducting an experiment and inject a drug into three mice. Thei
ID: 3068485 • Letter: S
Question
Suppose you are conducting an experiment and inject a drug into three mice. Their times for running a maze are 8, 10, and 15 seconds; the times for two control mice (i.e. not injected with the drug) are 5 and 9 seconds.Perform a two-sided permutation rest to determine if there is a statistically significant difference between mean maze completion time with the drug and without the drug. Suppose you are conducting an experiment and inject a drug into three mice. Their times for running a maze are 8, 10, and 15 seconds; the times for two control mice (i.e. not injected with the drug) are 5 and 9 seconds.
Perform a two-sided permutation rest to determine if there is a statistically significant difference between mean maze completion time with the drug and without the drug.
Perform a two-sided permutation rest to determine if there is a statistically significant difference between mean maze completion time with the drug and without the drug.
Explanation / Answer
We will perform two tailed t-test if the underlying distribution assumptions are assumed to be in place.
Null hypothesis: 1 - 2 = 0 or 1 = 2
Alternative hypothesis: 1 - 2 0
SE = sqrt[(s1^2/n1) + (s2^2/n2)]
= 2.89
Calculate Degrees of freedom (DF):
DF=(s1^2/n1 + s2^2/n2)^2 / { [ (s1^2 / n1)^2 / (n1 - 1) ] + [ (s2^2 / n2)^2 / (n2 - 1) ] }
= 2.74 (rounded to 3)
t = [ (x1 - x2) - d ] / SE
where x and x2 are two sample means and d is the difference between two means we want to perform statistical significant test for. In our case because we want to compare two means, this is 0.
hence t = (11-7)/2.89 = 1.39
Using the t-distibution we find out the probablity of P(t < -1.39) = .1294 and P(t > 1.39) =.1294
which is greater than the statistical significance level we assumed (.10). hence fail to reject the null hypothesis. which means that the drug effect cannot be ascertained at the significance level (10%).
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