5. The production control supervisor at a ish cannery in a the fill rates of can

Question

5. The production control supervisor at a ish cannery in a the fill rates of cans of tuna. The current canning process for 8 oz. cans of tuna has a standard deviation of 0.07 oz. Following changes to the canning process the production control supervisor sampled 30 (thirty) 8-oz cans of tuna and computed a standard deviation of 0.056 oz. Based on these sample results can the production changes have reduced the variability in the fill rate for 8oz. cans of tuna? U significance of 0.025 to conduct the hypothesis test. (20 points) supervisor conclude that the production process Pointsuna? Ho: HA: Critical Value and Decision Rule: Calculated Value: Conclusion:

Explanation / Answer

Given that,
population standard deviation ()=0.07
sample standard deviation (s) =0.056
sample size (n) = 30
we calculate,
population variance (^2) =0.0049
sample variance (s^2)=0.003136
null, Ho: =0.07
alternate, production process changes have reduced the variability, H1 : <0.07
level of significance, = 0.025
from standard normal table,left tailed ^2 /2 =45.722
since our test is left-tailed
reject Ho, if ^2 o < -45.722
we use test statistic chisquare ^2 =(n-1)*s^2/o^2
^2 cal=(30 - 1 ) * 0.003136 / 0.0049 = 29*0.003136/0.0049 = 18.56
| ^2 cal | =18.56
critical value
the value of |^2 | at los 0.025 with d.f (n-1)=29 is 45.722
we got | ^2| =18.56 & | ^2 | =45.722
make decision
hence value of | ^2 cal | < | ^2 | and here we do not reject Ho
^2 p_value =0.9321
ANSWERS
---------------
null, Ho: =0.07
alternate, H1 : <0.07
critical value: -45.722, reject Ho, if ^2 o < -45.722
test statistic: 18.56
decision: do not reject Ho
we don't have that production process reduced the variability

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