Question

XYZ Consumer Protection Authority (XYZ-CPA) is responsible for enforcing laws concerning weights of advertised essential items. XYZ-CPA routinely inspects packages to determine if the net weight of the contents of 20kg rice bags is at least as large as the net 20kg weight advertised on the bag.

A random sample of 96 bags were inspected where the net weight content of each of those bags was measured. The mean net weight content calculated from the sample is found to be 20.30kg with a standard deviation 1.20kg. Assuming that the net weight of 20kg rice bags is normally distributed, estimate with 95% confidence the actual mean net weight content of 20kg rice bags.

a) You were recently hired as a junior data analyst working for XYZ-CPA. Specify the appropriate formula you would use to solve the problem. Provide a brief reason why you chose the formula.

b) Obtain the 95% confidence interval estimate of the actual mean net weight content of 20kg rice bags. Display working.

c) Provide an interpretation of the answers you obtained in part b) in the context of the problem.

d) Present brief statements of two or three sentences to XYZ-CPA with regard to the outcome of the weight inspection or the quality control process that you just conducted. Should the XYZ-CPA be satisfied with the outcome? Yes or no? Why?

e) Assuming all of the other variables to do the calculation of confidence interval held constant, what would happen to the width of the interval when a higher sample size is used?

Answer 1

ū tt(0.975, 6-1) * Secù) Sol we have, no96 * In this problem we Sample mean n = 20.30kg Standard deviation of Sample S= 1.20 kg (a) The appropriate formula to find the 95% Cofidence interval of the mean is htt(0.975n-1+ Se (w) don't know the Population variance, so we we the t table Value as multiplier in margin of Essor- Also we have to find as) confidence interval that means we have a=0.05 so the table vake will be required at (1-4/2) level. * (1) The 9st confidence interval of mean is given by, att(0975,95) & s e a

= 20.309 1.98520 1-20 196 20.30 + 0.2431426 Lower limit= 20.06 upper limit= 20.54 () be interpreted as that the لیا The above ast. confidence interval can we can be ast. Confident mean net weight of tice bags is mos between 20.06kg and 20.54 kg. d) XYZ-CPA wanted to determine if the net weight of the contents of 20kg. bags is at least as lodge as the net 20kg weight advertised on the bag and from above 951. Confidence interval we can be ast. Sure that the mean net weight of rice bags than 20kg. So the XYZ-CPA should be satisfied with the outcome as the contents of rice bags is at least as large of the net 20kg. is more

e) In the formula of confidence interval, We can observe that the confidence interval is inversely proportional to the square root of the sample size. So, if the sample Size increases, the width of the interval become harrower. o please give me an upvote.