Question 135217
<b><font size = 7 color = "red">Stanbon's solution is incorrect. He divided alpha .10 by 2, getting .05, which should only have been done if the test had been two-tailed, but this is a right-tail test.   
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A coin was flipped 60 times and came up heads 38 times. (a) at the .10 level of significance, is the coin biased toward heads? Show your decision rule and calculations. 
<pre><b>
H<sub>0</sub>: p = {{{1/2}}}
H<sub>a</sub>: p > {{{1/2}}}

This is a one-tail (right-tail) test

z = {{{(pHAT-pO)/sqrt((pO(1-pO))/n)}}}

where {{{pHAT}}} = {{{38/60}}} = {{{19/30}}}
{{{pO=1/2}}}
{{{n}}} = {{{60}}}

z = {{{(19/30-1/2)/sqrt(((1/2)(1-1/2))/60)}}}

z = {{{(19/30-15/30)/sqrt(((1/2)(1/2))/60)}}}

z = {{{(4/30)/sqrt(((1/4))/60)}}}

z = {{{(2/15)/sqrt((1/4)(1/60))}}}

z = {{{(2/15)/sqrt(1/240)}}}

z = {{{(2/15)sqrt(240/1)}}}

z = {{{(2/15)sqrt(240)}}}

z = {{{(2/15)sqrt(16*15)}}}

z = {{{(2/15)4sqrt(15)}}}

z = {{{(8/15)sqrt(15)}}}

z = {{{2.065591118}}}

The .10 level of significance means that
we will reject the null hypothesis if 2.065591118
is greater than the z-value which has .10 (10%) of
area under the standard normal curve to the right 
of it.

Now there are two kinds of normal tables in use today,
and I don't know which you have.

If you have the kind that accumulates the area to the 
right of z=0, then subtract .5-.10 = .40, then look
through the body of the table and find the nearest
value to .4000, which is .3997, and observe that that
occurs at z = 1.28.

If you have the kind that accumulates the area from
the extreme left, then subtract 1-.10 = .90, then look
through the body of the table and find the nearest
value to .9000, which is .8997, and observe that that
occurs at z = 1.28.
 
Since our z-value of 2.065591118 is to the right of
1.28, then we reject the null hypothesis and conclude
that the probability of heads is greater than {{{1/2}}},
thus it is biased toward heads.

(b) Calculate a p-value and interpret it.

This is different way of completing the problem.  Instead
of finding the value of z which has .10 (10%) of the
area to the right of it, we find instead the area to the
right of the z value which we have calculated above, and 
then reject the null hypothesis only if that area is less 
than .10.

So we find the area to the right of

z = 2.065591118

So regardless of which table you have, look up the nearest
value of z to that, which is 2.07.  

If you have the kind of table that accumulates area from
z = 0, then you will read .4808, then subtract that from
.5000 to get the p-vale as .0192.

If you have the kind of table that accumulates area from
the extreme left, then you will read .9808, then subtract 
that from 1.0000 to get the p-value as .0192.  
 
Since this p-value is less than .10, our z-score has less
than .10, and we therefore reject the null hypothesis, and
thus we conclude that the coin is biased toward heads.

To interpret the p-value for this problem:

If the coin were not biased, then the probability of getting
38 or more heads out of 60 would be only .0192.  

Edwin</pre>