Homework Answers
20.
Given that,
population variance (σ^2) =25
sample size (n) = 20
sample variance (s^2)=36
null, Ho: σ^2 >=25
alternate, H1 : σ^2 <=25
level of significance, α = 0.1
from standard normal table,left tailed ᴪ^2 α/2 =27.204
since our test is left-tailed
reject Ho, if ᴪ^2 o < -27.204
we use test statistic chisquare ᴪ^2 =(n-1)*s^2/o^2
ᴪ^2 cal=(20 - 1 ) * 36 / 25 = 19*36/25 = 27.36
| ᴪ^2 cal | =27.36
critical value
the value of |ᴪ^2 α| at los 0.1 with d.f (n-1)=19 is 27.204
we got | ᴪ^2| =27.36 & | ᴪ^2 α | =27.204
make decision
hence value of | ᴪ^2 cal | > | ᴪ^2 α| and here we reject
Ho
ᴪ^2 p_value =0.0965
ANSWERS
---------------
null, Ho: σ^2 >=25
alternate, H1 : σ^2 <=25
test statistic: 27.36
critical value: -27.204
p-value:0.0965
decision: reject Ho
we have enough evidence to support the claim that the variance of
the sugar content of its yogurt is less than or equal to
25.
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