Question 1210349
<pre>
I decided to do a recently posted statistics problem.  

Not because I like them, I don't. But because high schools and colleges are
striving to make math courses practical in the workplace.  

Face it, algebra, geometry, and trig are no longer practical in any workplace,
for computers do it all.  Engineers these days do nothing but point and click.

Statistics is the only thing that's still practical in nearly every field.

So schools began to push statistics, because it only requires first year
algebra, and with computers doing all the substituting in formulas, even
less.

When high schools began pushing statistics, this site at first had a
deluge of statistics problems posted.  

However, no tutors on here would bother solving them, including yours truly, so
students learned this site was no good for posting stats problems, and this site
went to pot.  

So if you tutors want to see this site start thriving again, you'd better start
solving statistics problems. For that's the only way to salvage it.  

Face it, finding such things as integer solutions to randomly made-up equations
are as impractical in the workplace as making mud pies.

--------------------------

Anyway, here's my stab at a t-problem.  My stats knowledge is getting rusty.

Count these data.  There are 16.  This is less than (or equal to) 30. That's
considered a small sample.  So we will use the t-distribution, not the 
normal z-distribution.

20 14 23 22 22 15 21 18
23 17 14 25 15 21 20 14

Add them, get 304

Divide by 16, get {{{mean=xbar=19}}}

We want the standard deviation, s. 

{{{s}}}{{{""=""}}}{{{sqrt(sum((x[i]-xbar))^2/(n-1))}}}

Get the deviations from the mean, by subtracting 19 from each of those in the
list.

 1 -5  4  3  3 -4  2 -1
 4 -2 -5  6 -4  2  1 -5

Square the deviations from the mean:

 1 25 16  9  9 16  4  1
16  4 25 36 16  4  1 25

Add them, get 208 

Divide by n-1, or 16-1 or 15
Get {{{s^2=13.86666667}}}
{{{s=sqrt(s^2)=3.723797345}}}

Now the formula for the margin of error is 

{{{E = t*expr(s/sqrt(n))}}}

We calculate {{{s/sqrt(n)}}}{{{""=""}}}{{{3.723797345/sqrt(16)}}}{{{""=""}}}{{{3.723797345/4}}}{{{""=""}}}{{{0.93094593358}}}

Since this is a 2-tailed test, 90 percent confidence interval leaves 10 percent
on both ends, or half that, 0.05 on each end, giving the confidence interval of
90 percent in between these two critical values.  
   
So we look up the t-value for the number of 15 (1 less than the number in the
sample of 16) degrees of freedom for a 2-tailed test and find the value 1.753.

(or you can find the t-value on your TI-84 calculator by 

2nd VARS area:0.95, df:15 ENTER ENTER ENTER, 

and read more decimal values as 1.753050323)

{{{E = t*expr(s/sqrt(n))}}}{{{1.753050323*0.93094593358}}}{{{""=""}}}{{{1.63199507}}}

That's the margin of error.  

The confidence interval is {{{xbar +- matrix(1,3,margin,of,error)}}}

(19-1.63199507,19+1.6199507) = (17.36800493, 20.63199507).

If you use a table of t-values you won't get that many decimal places.

Edwin</pre>