Download Class 18 Lecture

Class 18,
June 23, 2014
Plan for Class 18:
1. The Problem of Hypothesis Testing (The case of Large
Samples)
Hypothesis Testing for Large Samples
Problem. Example 7.5, p.373.
X = {The amount of cornflakes discharged by a filling
machine into a standard box, in ounces}
The machine is design to discharge on average 12 ounces. A
statistical test below should decide whether there are
deviations from 12 ounces.
Test
H0: µx= 12
versus
H1: µx≠ 12
at the level of significance α=0.01.
First we construct the Rejection Region corresponding to
α=0.01.
These are numbers to the left of  z0.005  2.576 and to the
right of z0.005  2.576 .
Now, given sample we find (the book gives this data):
n  100, X  11.851, s  0.512 and calculate
X  0
  2.91.
s/ n
If Z gets int o the rejection region ( RR ), we reject H 0 in favor
the value of test statistic Z 
of the alternative hypothesis H a at the level of significan ce  .
Since indeed Z gets int o RR , we reject
H 0 in favor of the alternative hypothesis H a .
As we have shown in class, this procedure is justified by the
fact that the probability of type I error (we reject the correct
hypothesis) is α, a small number.
The above procedure can be restated (equivalently) by the
notion of the p-value.
p-value is the sum of tail areas corresponding to the points
X  0
s/ n
and

X  0
.
s/ n
In our case this will be to the left of

X  0
  2.91.
s/ n
X  0
  2.91 and
s/ n
to the right of
The sum of these two areas (the p-value) is
0.0036.
Once the p-value is found, we can easily say whether or not
we reject the null hypothesis for any level α. Namely,
if   p  value, we reject,
if   p  value, we don' t reject. .
Since our p-value is 0.0036, we can say for example that
for   0.0025 we don' t reject,
while for   0.005 we do reject.
Homework:
Read Example 7.5 on page 373.