)Will someone help me please and solve these two statistic problem for me?

As usual, folks complicate things to where their students cannot understand them. The standard error of the mean (which is just the standard deviation of the mean) is given by:

sd(pop)/sqrt(n) = 50 / sqrt(n)

This is a specific formula for when one assumes the population is infinite; The more general formula is

var(mean) = (var(pop)/n ) x { [N -n] / [N - 1]}

looking at the last term called a “finite population correction factor“:

N - n

------

N - 1

one can see two things:

1. as n, the sample size, approaches N, the population size, the error reduces until at the limit it is zero.

2. As N, the population size increases, the correction eventually becomes "1" at N = infinity. Consequently, we normally do not even consider the finite population correction in most cases as our populations are large.

so, lets look at your problem:

N = 60000

n = 15000

sd = 50

std err = 50 / sqrt(15000) = 50 / 122 = .41

that assumes infinite an population. now, we correct

N - n / N - 1 is the correction factor.

(60000 - 15000) / (60000 - 1) = 45000/59999 = .75

since this corrects the VARIANCE, we take the sqrt to correct the s.d.: sqrt(.75) = .86

and our final answer is .41 x .86 = .35

you should be able to do the rest.

always,

tony

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