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<h2>Answer with explanation:</h2>
Let be the average weight of chocolate chips in each bag.
As per given , we have
Since is left-tailed , so we perform a left-tailed test.
Also, the population standard deviation is known .
So we use z-test.
Test statistic :
Substitute given value ,
Significance level = 0.05
Decision rule : Reject when p-value < 0.05.
By z-table , P- value for left-tailed test = P(z<-1.96)= 1- P(z<1.96)
[∵ P(Z<-z)=1-P(Z<z)]
= 1- 0.9750
=0.025
Since , P-value(0.025) < 0.05 , so we reject the null hypothesis.
We support the claim that the machine is underfilling the bags at 0.05 significance level .