John oversees a bottle-filling machine in a company. The amount of fluid dispens
ID: 3330654 • Letter: J
Question
John oversees a bottle-filling machine in a company. The amount of fluid dispensed into each bottle is approximately normally distrbuted with an unknown population standard deviation. On a particular day, a random sample of 50 bottles yielded a mean of 357.2 ml and a standard deviation of 15.8 ml John then concluded that the population standard deviation of the fluid dispense amount by the machine is greater than 15 ml. 17. The lead engineer wants to use a 0.01 significance level to test John's claim. (a) Identify the null hypothesis and alternative hypothesis. termine the test statistic. Show all work writtng the correct test statistic, without supporting work will recerve no credit. () Determine the P-value for this test Show all work writtng the correct Pavaiue, without supporting work, will recetve no credit (d) Is there sufficient evidence to support John's claim that the population standard deviation of the fluid dispense amount by the machine is greater than 15 ml? Explaan.Explanation / Answer
Given that,
population standard deviation ()=15
sample standard deviation (s) =15.8
sample size (n) = 50
we calculate,
population variance (^2) =225
sample variance (s^2)=249.64
null, Ho: <225
alternate, H1 : >225
level of significance, = 0.01
from standard normal table,right tailed ^2 /2 =74.919
since our test is right-tailed
we use test statistic chisquare ^2 =(n-1)*s^2/o^2
^2 cal=(50 - 1 ) * 249.64 / 225 = 49*249.64/225 = 54.37
| ^2 cal | =54.37
critical value
the value of |^2 | at los 0.01 with d.f (n-1)=49 is 74.919
we got | ^2| =54.37 & | ^2 | =74.919
make decision
hence value of | ^2 cal | < | ^2 | and here we do not reject Ho
^2 p_value =0.2774
ANSWERS
---------------
null, Ho: <225
alternate, H1 : >225
test statistic: 54.37
critical value: 74.919
p-value:0.2774
decision: do not reject Ho
we don't have evidence that fluied dispense amount by the machine is greater than 15 ml
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