2
$\begingroup$

The rope used in a lift produced by a certain manufacturer is known to have a mean tensile breaking strength of 1700 kg and standard deviation 10.5kg. A new component is added to the material which will, it is claimed, decrease the standard deviation without altering the tensile strength. Random samples of 20 pieces of the new rope are tested to destruction and the tensile strength of each piece is noted. The results are used to calculate unbiased estimates of the mean strength and standard deviation of the population of new rope. These were found to be 1724 kg and 8.5kg.

(i) Test, at the 5% level, whether or not the variance has been reduced.

(ii) What recommendation would you make to the manufacturer?

my answer*

i want to simply know when using hypothesis is it possible to use

NULL HYPOTHESIS : population variance(sigma^2) = 10.5^2

ALTERNATE HYPOTHESIS : variance has reduced ==>population variance < 10.5^2

then test statistic =chi square = (n-1)s^2/(sigma)^2 = (20-1)8.5^2/10.5^ = 12.4512

degree of freedom =20-1=19

i have a problem with choosing the critical chi square value

i have chosen chi square critical= chi square for .05 sl and df of 19 =10.1

Since test statistic doesnt fall in region of rejection we don't reject Null hypothesis. hence Variance has not been reduced. please help me with this problem

$\endgroup$
1
$\begingroup$

No sure what you mean by "sl". You want a number $c$ such that $\Pr(\chi^2_{19}<c)=0.05$. If the test statistic is less than that, then you reject $H_0$; otherwise you don't. (The software I'm using gives $c=10.11701$.)

You're OK, except that as your bottom line, instead of concluding that the variance has not been reduced, you should conclude that you haven't shown that it has been reduced. That could be because you need a more sensitive test to detect a small reduction, and that more sensitive test could come from just using a larger sample.

$\endgroup$
  • $\begingroup$ thank you for the answer!! sl= significance level :-) $\endgroup$ – ChampR May 25 '13 at 16:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.