Statistics questions sample mean 17.55 standard deviation 7.99 t value 1.440

label Statistics
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Mar 9th, 2015

We have a sample size of n = 7, so 6 degrees of freedom. We should use the t-distribution at 0.9 with df = 6:

t = t.inv(p=0.9, df=6) = 1.440


We also have a standard error = sample standard deviation / sqrt (sample size)

= 7.99/sqrt(7)

= 3.02


So the lower and upper values for the 80% confidence interval are given by:

mean +/- t_crit x SEM

= 17.55 +/- 1.440 x 3.02

= 17.55 +/- 4.348

= 13.20 to 21.90



Mar 9th, 2015

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Mar 9th, 2015
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