Hi again, wta2k!
I see you're still working on hypothesis testing, a very intersting
subject.
According to how the question is stated, it's not actually an
hypothesis testing of two sample means or proportions; it's rather a
testing of two sample variances. What you want to check is wether the
two variances are equal or one of them is bigger than the the other
one. As you will see, this can be easily done.
For the remainder of the question, let us assume that both sipping
times are normally distributed, one with mean 3.4 and the other one
with mean 3.6. Anyway, as the questions states, the mean is not
important. The assumption of a normal distribution greatly simplifies
the analysis. Hypothesis testing of variance under non-normal
distributions is not easy at all.
Let's call 'S' the sample variance of a normal distribution, 's' the
(unkown) population variance, and 'n' the sample size (number of
observations). It can be shown that (n-1)S/s follows a chi-square
distribution with n-1 degrees of freedom.
Here we have two samples. Let us call S1 and S2 the sample variances
of group 1 and 2 respectively (the 2 suppliers), n1 and n2 their
sample sizes, and s1 and s2 the population variances. We have thus two
chi-square distributions:
(n1-1)S1/s1 is a chi-square distribution with n1-1 degrees of freedom,
and
(n2-1)S2/s2 is a chi-square distribution with n2-1 degrees of freedom.
Finally, it can be shown (this the last theoretical part of the
answer, right after this I get right to your specific question) that
the ratio of two chi-square random variables divided by their
respective degrees of freedom follow and F distribution
F-Distribution
http://mathworld.wolfram.com/F-Distribution.html
(just read the first paragraph and formula)
The F Distribution has two parameters: "degrees of freedom (df) of the
numerator" (that's the df of the first chi-sqare random variable) and
df of the denominator (the df of the second one).
Why is this relevant to your question? You want to test the null
hypothesis that both variances are equal versus thh alternative
hypothesis that the variance of group 2 is bigger than the variance of
group 1. The statistic we will use here will be S2/S1. Why? First, it
makes sense for what you want to test. Under the null hypothesis, this
ratio should be close to 1. We would reject the null hypothesis if
this ratio is "sufficiently" larger than 1 (this "sufficiently" will
depend on the level of significance of the test). Second, this
statistic is convenient because, under the null hypothesis that s1=s2
(notice that this is 's', the population variance, and not 'S'), it
follows an F distribution. Proof of this, if you need it, is after the
question.
So, we need to set the rejection area. We are looking for k such that
Prob ( S2/S1 > k ) = 0.025
Then we will check if S2/S1 is actually bigger or smaller than k. If
it is bigger, we will reject the null hypothesis, because S2/S1 will
be "sufficiently" bigger than 1. If it's not, we can't reject the null
hypothesis.
We know that S2/S1 follows an F distribution. It has 29 numerator df
(because group 2 has 30 observations) and 23 denominator df. Now, you
can check any F distribution table. For example, go to
F-distribution tables
http://www.stat.ucla.edu/~dinov/courses_students.dir/Applets.dir/F_Table.html
Now go to the table with alpha=0.025 (your level of significance). Go
to column '30' (as this is the number closest to 29 in this table) and
row '23'. You get the value 2.239.
So, should you reject the null hypothesis that the population
variances are equal? No, because S2/S1 = 0.7/0.4 = 1.75 < 2.239. The
ratio between sample variances was not big enough for us to conclude
the the population variance of group 2 is indeed bigger than group
one's. We can't reject the null hypothesis that s1=s2. The fact that
the sample variance of group 2 was bigger is most probably chance.
The steps we did here in order to come to this conclusion were:
1. Determine that the ratio between the sample variances follow and F
distribution with numerator df equal to the first group and
denominator df equal to the second group. In this case it was en F
distribution with 29 numerator df and 23 denominator df.
2. Determine the rejection area, by checking in a table what's the
F-value that makes the probability of the previous ratio being greater
than this F-value is equal to the level of significance (0.025 in this
case).
3. Determine wether the actual ratio S2/S1 is greater or smaller than
the F-value you found above.
For more information on this subject, you might also want to check the
following site
Test of significance for Two Population Variances
http://home.xnet.com/~fidler/triton/math/review/mat170/fdist/fdist1.htm
I hope this was clear enough for you. If you have any further
questions, please request a clarificiaton before rating my answer.
Best regards,
elmarto
--------
In case you're interested, here's the proof that S2/S1 follows an
F-distribution as described above under the null hypothesis that
s1=s2. We had that the ratio of two variables with chi-sqare
distributions, divided by their degrees of freedom, had an
F-distribution. Then
( (n1-1)S1/s1 ) / (n-1)
_______________________
( (n2-1)S2/s2 ) / (n-2)
follows an F- distribution. But, under the null hypothesis, the terms
s1 and s2 cancel out. So do the terms (n1-1) and (n2-1). So we are
only left with S1/S2. |
