Given that,
population standard deviation (sigma)=0.31114
sample standard deviation (s) =0.0548
sample size (n) = 5
we calculate,
population variance (sigma^2) =0.0968080996
sample variance (s^2)=0.00300304
null, Ho: sigma =0.31114
alternate, H1 : sigma !=0.31114
level of significance, alpha = 0.05
from standard normal table, two tailed chisqr^2 alpha/2
=9.488
since our test is two-tailed
reject Ho, if chisqr^2 o < - OR if chisqr^2 o > 9.488
we use test statistic chisquare chisqr^2 =(n-1)*s^2/o^2
chisqr^2 cal=(5 - 1 ) * 0.00300304 / 0.0968080996 =
4*0.00300304/0.0968080996 = 0.12
| chisqr^2 cal | =0.12
critical value
the value of |chisqr^2 alpha| at los 0.05 with d.f (n-1)=4 is
9.488
we got | chisqr^2| =0.12 & | chisqr^2 alpha | =9.488
make decision
hence value of | chisqr^2 cal | < | chisqr^2 alpha | and here we
do not reject Ho
chisqr^2 p_value =0.9983
ANSWERS
---------------
a.
standard deviation 1=0.0548
standard deviation 2 =0.3114
null, Ho: sigma =0.31114
alternate, H1 : sigma !=0.31114
b.
test statistic: 0.12
critical value: -9.488 , 9.488
p-value:0.9983
c.
decision: do not reject Ho
option: No
we do not have enough evidence to support the claim that difference
of standard deviations.
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