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2 answers:
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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 .
Answer:
The number of weekday hours are 5.
Step-by-step explanation:
<em>Let the number of weekday hours be "x".</em>
<em>The total week hours are 26, then number of weekend hours is (26-x) .</em>
<em>The charges for weekdays are $2 per hour and for weekends $10 per hour.</em>
The total money spent is $220.
x = 5.
Thus weekday hours utilized are 5 hours, and weekend are (26-5) 21 hours.