3. This is a \"quality control\" problem. You manage several grocery stores and

Question

3. This is a "quality control" problem. You manage several grocery stores and are concerned that one of them may be failing to monitor their produce properly. This grocery store sells 5-pound bags of whole potatoes. You know that there will be variability; some bags will weigh more than others. You also know that you will lose profits if the bags weigh too much, but it would be illegal if the bags weighed too little The mean across all stores is 5 pounds and the population standard deviation is 1 pound. So you takea sample of 16 bags of potatoes from that store and you find the average weight of that sample is 6 pounds. Please create a 95% confidence interval to estimate whether this store is over or under-filling their bags of potatoes a) Would you use a z-score or a t-score for this confidence interval? Hint: do you know the population variability?) b) What would your critical z-score be? c) What would your standard error of the mean be?_ d) Given that the average weight from our sample is 6 pounds and the standard error of the mean of 0.25: Calculate the lower boundary of your confidence interval: Calculate the upper boundary of your confidence interval: e) Does the "ideal weight" of 5 pounds fall within that range? What advice would you give the manager of that particular store? g) You would use a z score of to construct a 95% confidence interval, and a z-score to construct a 99% confidence interval h) What is the point estimate for of the population mean?

Explanation / Answer

Answering 4 parts as per Chegg policies

A) Since I am aware of the population variance I use z score.

B) critical z = +-z(a/2) = 1.96

C) SE = sd/sqrt(n) = (1/sqrt(16)) = 0.25

D) lower = mean -z(a/2)*SE = 6-1.96*0.25 = 5.51

upper = mean +z(a/2)*SE = 6+1.96*0.25 = 6.49

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.