Question

Suppose that you are wanting to estimate the mean of a population in which the standard deviation, $\sigma$ , is known to be 5. You take a random sample of size n=50 from the population and determine that the mean of the sample is 22.4.

a. What is the standard error (of the mean) to two decimal places?

b. If you wanted 94% confidence in your results, then what would be the maximum allowable error (error of the estimate), E, to two decimal places?

c. In general, if you wanted to estimate the mean of a population with standard deviation = $\sigma$ , how would the standard error be affected if quadrupled the size of your random sample? That is, by choosing samples of size 4n instead of n, the standard error would: (double, be cut in half, Quadruple, be only a quarter of its original, be unaffected)

Answer 1

a)

std error=σx̅=σ/√n=5/√50 =

0.71

b)

for 94% confidence ; critical z =1.88

therefore maximum allowable error E =z*standard error =1.88*0.71 =1.33

c)

be cut in half

(since standard error is inversely proportional to square root of sample size)