Homework Answers
Given that,
population standard deviation (σ)=15
sample standard deviation (s) =11
sample size (n) = 22
we calculate,
population variance (σ^2) =225
sample variance (s^2)=121
null, Ho: σ =15
alternate, H1 : σ !=15
level of significance, α = 0.05
from standard normal table, two tailed ᴪ^2 α/2 =32.671
since our test is two-tailed
reject Ho, if ᴪ^2 o < - OR if ᴪ^2 o > 32.671
we use test statistic chisquare ᴪ^2 =(n-1)*s^2/o^2
ᴪ^2 cal=(22 - 1 ) * 121 / 225 = 21*121/225 = 11.293
| ᴪ^2 cal | =11.293
critical value
the value of |ᴪ^2 α| at los 0.05 with d.f (n-1)=21 is 32.671
we got | ᴪ^2| =11.293 & | ᴪ^2 α | =32.671
make decision
hence value of | ᴪ^2 cal | < | ᴪ^2 α | and here we do not reject
Ho
ᴪ^2 p_value =0.9568
ANSWERS
---------------
null, Ho: σ =15
alternate, H1 : σ !=15
test statistic: 11.293
critical value: -32.671 , 32.671
p-value:0.9568
decision: do not reject Ho
we do not have enough evidence to support the claim that standard
deviation of number of people interviewed by the corporation in one
month differs from 15.
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