Question

Suppose for a population, ? = 45, ? = 8. A sample of size n = 31 is selected.

a) What is the standard deviation of sample mean? i.e. ?([?(x)]) =

b) What is the z-score of [?(x)] = 41.5?

c)If the sample standard deviation is s = 9, what is the standard error of sample mean? i.e. s.e.([?(x)]) =

d)If the sample standard deviation is s = 9, what is the t-score of [?(x)] = 41.5?

Answer 1

Solution:

Given that,

= 45

= 8

n = 31

So,

a ) The standard deviation of sample mean is  

   =  ( /n) = (8/ 31 ) = 1.4368

The standard deviation of sample mean = 1.4368

b ) X = 41.5

Using z-score formula,

Z = X -   /

Z = 41.5 - 45 / 8

Z = -3.5 / 8

Z = - 0.44

c ) S = 9  

The standard error of sample mean is S

S = (s /n)

S = 9 / 31

S = 1.6164

The standard error of sample mean = 1.6164

d ) S = 9

The t-score is - / S

   - / S = 41.5 - 45 / 9

= -3.5 / 9

= - 0.39

The t-score = - 0.39