13.60
In past studies, a fast-food franchise has found the standard
deviation of weights for quarter-pound hamburgers to be approximately
0.28 ounces after cooking. A consultant has suggested a new procedure
for hamburger preparation and cooking that she believes will reduce
the variability in the cooked weight of the hamburgers. In a test of
the new method, 61 hamburgers are prepared, cooked, and weights, with
the resulting cooked weights listed in data file XR13060. At the 0.05
level, and assuming a normal distribution for the weights, examine the
possibility that the new procedure is actually no better than the old
one. (Chi-Square Application)

Hi, I need clear step by step work answers. However, not long
explanations or professional answers. Thank you very much.

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Request for Question Clarification by websearcher-ga on 02 Apr 2003 21:16 PST

Hi:

You haven't supplied file XR13060. 

websearcher-ga

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Clarification of Question by wta2k-ga on 03 Apr 2003 09:05 PST

Sorry,... This is the information of the data file "XR13060". (total,
from 01 to 62)

Ounces
4.155
3.707
3.674
4.458
3.915
4.034
3.851
4.562
4.000
3.949
3.956
3.577
4.015
4.610
4.020
3.983
3.964
4.003
3.856
4.019
3.587
3.902
3.952
3.907
4.256
4.235
3.757
3.903
4.166
4.037
4.132
3.914
4.007
4.368
3.822
4.072
4.276
4.307
3.822
4.270
3.978
4.311
4.235
4.354
4.104
3.917
4.039
3.961
4.079
3.807
4.129
3.984
4.022
4.164
3.968
3.652
3.634
4.198
4.108
4.065
4.180

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Clarification of Question by wta2k-ga on 03 Apr 2003 09:07 PST

... from 01 to 61, not 62.

Hello wta2k!
I have found a page that shows the procedure needed in order to answer
your question. You might want to check it first, then I will provide
the answer to your particular exercise

Hypothesis testing about variances
http://wind.cc.whecn.edu/~pwildman/statnew/section_6_hypothesis_testing_about_variances.htm

Now, regarding your exercise. First of all you need to compute the
sample variance from the data you provided. The formula for sample
variance is in the following page

Sample Varuance
http://mathworld.wolfram.com/SampleVariance.html
(it's equation (3) )

I just used Excel to compute it, finding the the sample variance for
your data (let's call it s^2) is 0.0484. The variance before the
adjustment of the machine is 0.28^2 (0.28 squared) = 0.0784. So you
want to test wether 0.0484 is significantly lower than 0.0784. I'll
assume then that you want to do a one-tail test: the null hypothesis
is that the variance is equal or higher than 0.0784; the alternative
hypothesis is that it is lower than 0.0784 (that is, that the new
procedure in making hamburgers really reduces variability in weight)

Then, you have to find Prob(s^2 =< 0.0484) given that the population
variance is 0.0784. If the probability you find here is less than
0.05, you will reject the null hypothesis: it's very unlikely that you
observe a sample variance of 0.0484 or less with a population variance
of 0.0784.

In order to compute this probability, you need to know that
s^2*(n-1)/sigma^2 follows a chi-square distribution with n-1 degrees
of freedom, n being the number of observations (61 in this case), and
sigma^2 being the population variance (0.0784 in this case). So,

 Prob( s^2 =< 0.0484 )
=Prob( s^2*60/0.0784 =< 0.0484*60/0.0784 )
=Prob( chi =< 37.04)
(I write chi instead of s^2*60/0.0784 because that expression follows
a chi-square distribution)

Now, you can just look up in a table the probability you're looking
for. The following link gives the chi-square table.

Table: Chi-Square Probabilities
http://www.richland.cc.il.us/james/lecture/m170/tbl-chi.html
(df stands for degrees of freedom; in this case df=61-1=60)

This table gives the probability that a chi-square is *greater* than
the number you provide, so you just have to substract the probability
found here from 1. Look the table, go to the line where df=60, then
find the number that is closest to 37.04. In this table, that number
is 37.485. Now go all the way up through that column to find that the
probability of a chi-square being greater than that number is 0.99.
Therefore, the probability you're looking is 1-0.99=0.01.

Since 0.01 is smaller than the level of significance set for this test
(0.05) we REJECT the null hypothesis: it's "unlikely" that we observe
the sample variance we observed if the population variance were
0.0784. The conclusion is that, at the 0.05 level of significance, the
new procedure was successful at reducing the variability in the weight
of hamburgers.

I hope the answer was clear enough. If you have any further questions,
please request a clarification before rating my answer, I will be more
than happy to provide further assistance to you in this matter.


Best regards,
elmarto

Thank you for your quick help!!