Answer:
Bags are being underfilled by the machine
Explanation:
Solution:-
- The potato chips manufacturer wants to statistically determine whether the machine is operating correctly based on its weight control.
- The weight control limit was placed on the machine for 404.0 grams.
- He believes that the machine is underfilled.
- So, he takes a sample of size n = 48 bags and weigh them separately and get the following data for his sample.
Sample Mean ( x^ ) = 400.0 grams
Variance ( σ ) = 169 grams
- He wants to statistically check his doubts on the machine operability at a significance level ( α ) of 0.02.
- We will make an hypothesis that the machine works fine. So,
Null hypothesis: u = 404.0 grams
- We will test against what he believes to be the case i.e " under-filling of bags "
Alternate hypothesis: u < 404.0 garms
- Next we will determine what test is to be applied in this case.
- Population variance ( σ^2 ) is known
- Sample size n > 30
- For the above two conditions we can apply the use standard normal distribution ( Z-score ).
- We are testing for underfilling of bags. So we will apply one-left tail test.
- Hence, the critical value for the Alternate hypothesis to hold true would be:
P ( Z < Z-critical ) = α
P ( Z < Z-critical ) = 0.02
Z-critical = -2.06
- Next we compute the Z-statistics associated with the sample data obtained:
Z-test = ( x^ - u ) / √ ( σ^2 / n )
Z-test = ( 400 - 404 ) / √ ( 169 / 48 )
Z-test = -4 / 1.87638
Z-test = -2.13176
- For the left-one tailed test the Z-critical value provide the limit underneath which all the statistical test values lie in the rejection region.
Therefore,
-2.13176 < -2.06
Z-test < Z-critical. ( Null hypothesis Rejected )
Conclusion:
- The statistics value lies in the rejection region; hence, the Null hypothesis is rejected. With can conclude with 98% confidence that the claim made on bags being underfilled by the machine is evident.