Standard Deviation?

A simple random sample of six male patients over the age of 65 is being used in a blood pressure study. The standard error of the mean blood pressure of these six men was 22.8. What is the standard deviation of these six blood pressure measurements?





a. 9.31


b. 50.98


c. 55.85


d. 136.8

Standard Deviation?
the answer is A, 9.31








Let X1, X2, X3, ... , Xn be a random sample. The mean, xbar is found by:





xbar =





n


∑ Xi / n


i = 1





the sample variance is:





n


∑ (Xi - xbar)² / (n - 1)


i = 1





This can be written in another form that is easier to compute





1/(n-1) * { [ ∑ (Xi)² ] - n * xbar² }





the variance is divided by n -1, not n. this is done because if you divide by n you have a biased estimator for the population variance. using n - 1 yields an unbiased estimator.





the sample standard deviation is found by taking the square root of the variance.








variance of the sample here is 22.8 ^ 2





the standard deviation of the mean of the six values is:





sqrt( 22.8^2 / 6 ) = 9.31