Question 310647
<pre><b>


_
x = 61.7, {{{sigma}}} = 14

{{{H[0]}}}: {{{mu=65}}}
{{{H[a]}}}: {{{mu<>65}}}


{{{z=(xBAR-mu)/(sigma/sqrt(n))=(61.7-65)/(14/sqrt(30))=-1.29}}}

No, we cannot reject the claim at {{{alpha=0.05}}} because z 
is neither less than -1.96 nor greater than 1.96.

We could also do it by computing the p-score, the probability
that xBAR would be at least as far off from 65 as 61.7 if the mean 
were actually 65.     

To find that p-score we find the area to the left of the z-score -1.29
as .5-.4015=.0985 and double it since this is a two-tail test.  That
gives the p-score of .197.

So we would only reject the null hypothesis if it were that {{{alpha>.197}}}.
But {{{alpha}}} is lower that that, only .05.

Edwin</pre>