please help with both, thank you very much!
Solution-1:
Ho:sigma=29
Ha:sigma<29
For the given sample ,n=9
sample standard deviation=7.53326
F=(n-1)*s^2/sigma^2
=(9-1)* 7.53326^2/29^2
=0.5398336
chi sq critcal value
=CHISQ.INV(0.05,8)
=2.732636793
answer:
test statistic is less than the critical value.Therefore,we cannot say that supermodel's weights are less variable than the general popualtion
OPTION C
Solution-2:
Ho:indpendent
Ha:dependent
chi sq=2.161
chi sq critical vale
=CHISQ.INV.RT(0.05,2)
=5.991464547
p=0.3394
ANSWER:
There is not evidence to reject the claim that the high school and section of participants are independent because
the test value 2.161<5.991
OPTION B
please help with both, thank you very much! Assume that the weights of adult women are...
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