Question 588065
<pre>
{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, exp(-x^2/2)), 
 
green(line(-1.645,0,-1.645,exp(-1.645^2/2)),line(1.645,0,1.645,exp(-1.645^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z)
  
)}}}

The left-most green line is erected where z=-1.645. 

The right-most green line is erected where z=+1.645. 

The area bounded by the curve, the z-axis, and those two green lines is 90% of
the total area between the curve and the z-axis.  The area to the left of the
left green line is 5% of the total area, and the area to the right of the right
green line is the other 5% of the total area between the curve and the z-axis.

Your are testing the hypothesis 

H<sub>0</sub>:  <font face = "symbol">m</font> = 500

against the alternative hypothess

H<sub>a</sub>:  <font face = "symbol">m</font> &#8800; 500
 
Suppose you took a sample of size n=100 and 
found that the sample mean was &#11571; = 493 and
the standard deviation were s = 50

Then you would calculate this test statistic:

     &#11571; - <font face = "symbol">m</font> 
z = 覧覧覧覧
       {{{s/sqrt(n)}}}


      493 - 500 
z = 覧覧覧覧覧覧 = -1.4
         {{{50/sqrt(100)}}}

This would correcpond to the red line drawn
at -1.4

{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, exp(-x^2/2)), 

red(line(-1.4,0,-1.4,exp(-1.4^2/2))),



green(line(-1.645,0,-1.645,exp(-1.645^2/2)),line(1.645,0,1.645,exp(-1.645^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z)

)}}}

and since the 1.4 value of z is between the
two green bars, the mean of 493 is close
enough to 500 with a sample that size with that
standard deviation, and so we would fail to
reject the hypothesis.

However, suppose you took a sample of size n=64 and 
found that the sample mean was &#11571; = 495 and
the standard deviation were s = 20
Then you would calculate this test statistic:

     &#11571; - <font face = "symbol">m</font> 
z = 覧覧覧覧
       {{{s/sqrt(n)}}}

      495 - 500 
z = 覧覧覧覧覧覧 = -2.0
         {{{20/sqrt(64)}}}
This would correcpond to the red line drawn
at -2.0
{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, exp(-x^2/2)), 
red(line(-2.0,0,-2.0,exp(-2.0^2/2))),
green(line(-1.645,0,-1.645,exp(-1.645^2/2)),line(1.645,0,1.645,exp(-1.645^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z)
)}}}
and since the 2.0 value of z is NOT between the two green bars, the mean of 495
with a sample that size with that standard deviation, is not close enough to
500, and so we would reject the null hypothesis.

Notice that a lot depends on the size and the standard deviation of the sample we take.

Now since this is a 2-tail test, the sample mean could be larger than 500

Suppose you took a sample of size n=81 and 
found that the sample mean was &#11571; = 510 and
the standard deviation were s = 70
Then you would calculate this test statistic:

     &#11571; - <font face = "symbol">m</font> 
z = 覧覧覧覧
       {{{s/sqrt(n)}}}

      510 - 500 
z = 覧覧覧覧覧覧 = -1.29
         {{{50/sqrt(100)}}}

This would correcpond to the red line drawn
at -1.29
{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, exp(-x^2/2)), 
red(line(1.29,0,1.29,exp(-1.29^2/2))),
green(line(-1.645,0,-1.645,exp(-1.645^2/2)),line(1.645,0,1.645,exp(-1.645^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z)
)}}}
and since the 1.29 value of z is between the
two green bars, the mean of 510 is close
enough to 500 with a sample that size with that
standard deviation, and so we would fail to
reject the hypothesis.

However, suppose you took a sample of size n=150 and 
found that the sample mean was &#11571; = 506 and
the standard deviation were s = 40
Then you would calculate this test statistic:

     &#11571; - <font face = "symbol">m</font> 
z = 覧覧覧覧
       {{{s/sqrt(n)}}}

      506 - 500 
z = 覧覧覧覧覧覧 = 1.84
         {{{40/sqrt(150)}}}

This would correcpond to the red line drawn
at +1.84
{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, exp(-x^2/2)), 
red(line(1.84,0,1.84,exp(-1.84^2/2))),
green(line(-1.645,0,-1.645,exp(-1.645^2/2)),line(1.645,0,1.645,exp(-1.645^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z)
)}}}
and since the 1.84 value of z is NOT between the two green bars, the mean of 506
with a sample that size with that standard deviation, is not close enough to
500, and so we would reject the null hypothesis.

Edwin</pre>