If 10 determinations of specific heat of iron had a standard
deviation of 0.0086, test the null hypothesis that σ=0.01 for such
determinations. Use the alternative hypothesis σ<0.01 and level
of significance 0.05.
given data are:-
sample size (n) =10
sample sd (s) = 0.0086
hypothesis:-
test statistic is:-
df = (n-1) = (10-1) = 9
p value :-
(as this is a left tailed test)
[in any blank cell of excel type CHISQ.DIST(6.6564,9,TRUE) ]
decision:-
p value = >0.05 (alpha)
so. we fail to reject the null hypothesis at 0.05 level of significance.
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