Question

a wine producer intends to limit the variance in the amount of alcohol of a bottle to be less than .15 to ensure quality control. A random sample of 45 bottles resulted in a variance of .138. Use a =.05 to test if the batch of lines should pass the quality control

Answer 1

Given that,
population variance (σ^2) =0.15
sample size (n) = 45
sample variance (s^2)=0.138
null, Ho: σ^2 =0.15
alternate, H1 : σ^2 <0.15
level of significance, α = 0.05
from standard normal table,left tailed ᴪ^2 α/2 =60.481
since our test is left-tailed
reject Ho, if ᴪ^2 o < -60.481
we use test statistic chisquare ᴪ^2 =(n-1)*s^2/o^2
ᴪ^2 cal=(45 - 1 ) * 0.138 / 0.15 = 44*0.138/0.15 = 40.48
| ᴪ^2 cal | =40.48
critical value
the value of |ᴪ^2 α| at los 0.05 with d.f (n-1)=44 is 60.481
we got | ᴪ^2| =40.48 & | ᴪ^2 α | =60.481
make decision
hence value of | ᴪ^2 cal | < | ᴪ^2 α | and here we do not reject Ho
ᴪ^2 p_value =0.6233
ANSWERS
---------------
null, Ho: σ^2 =0.15
alternate, H1 : σ^2 <0.15
test statistic: 40.48
critical value: -60.481
p-value:0.6233
decision: do not reject Ho
we do not have enough evidence to support the claim that a wine producer intends to limit the variance in the amount of alcohol of a bottle to be less than .15 to ensure quality control