Nylon bars were tested for brittleness (Bennett and Franklin 1954). Each of 280 bars was molded under similar conditions and was tested in five places. Assuming that each bar has uniform composition, the number of breaks on a given bar should be binomially distributed with five trials and an unknown probability p of failure. If the bars are all of
1
the same uniform strength, p should be the same for all of them; if they are of different strengths, p should vary from bar to bar. Thus, the null hypothesis is that the p's are all equal. The following table summarizes the outcome of the experiment:
\table[[Breaks/Bar,Frequency],[0,157],[
x^(˙)
,69],[
3
,35],[4,17],[5,1]]
a. Under the given assumption, the data in the table consist of 280 observations of independent binomial random variables. Find the mle of
p
.
b. Test the agreement of the observed frequency distribution with the binomial distribution using Pearson's chi-square test using
5%
significance. Note: The general form is
\chi ^(2)=\sum_(i=1)^m (|O_(0)-E_(i)|^(2))/(E_(i))
, so think about what the observed and expected counts are for each possible value of
x
, if we assume that each bar has a binomial distribution with
n=5
and common
p
.
Hint: Here, you can look at each possible
x
value, and how many you would expect (under
H_(0)
) to observe with that possible
x
value.
c. Let
x_(i)∼bin(n_(i),p_(i))
, for
i=1,dots,m
, be independent. Derive a likelihood ratio test for the hypothesis
H_(0):p_(1)=p_(2)=cdots=p_(m)
against the alternative hypothesis that the
p_(i)
are not all equal. Show that the test statistic is equivalent to
\sum_(i=1)^m ((x_(i)cdotsn_(i),(hat(p)))^(2))/(n_(i)(tilde(p))(1-(hat(p))))
. What is the large sample distribution of the test statistic?
Hint: Use the Taylor Series Expansion similar to how we did in class, taking
x=hat(p)_(4)
and
x_(0)=hat(p)
.
d. Use the test you derived in c to test the relevant hypothesis for this data. Does it agree with your test in part b?
Hint: Here, it is a little different than before. You are treating each bar as an individual binomial random variable, and considering what you expect (under
H_(0)
) the outcome to be for EACH rv.ONLY C AND D