A machine that is programmed to package 2.65 pounds of cereal is being tested fo

Question

A machine that is programmed to package 2.65 pounds of cereal is being tested for its accuracy. In a sample of 49 cereal boxes, the sample mean filling weight is calculated as 2.65 pounds. The population standard deviation is known to be 0.06 pound. Use Table 1. a-1. Identify the relevant parameter of interest for these quantitative data. The parameter of interest is the average filling weight of all cereal packages. The parameter of interest is the proportion filling weight of all cereal packages. a-2. Compute its point estimate as well as the margin of error with 99% confidence. (Round intermediate calculations to 4 decimal places. Round "z-value" to 3 decimal places and final answers to 2 decimal places.) Point estimate Margin of error b-1. Calculate the 99% confidence interval. (Use rounded margin of error. Round your final answers to 2 decimal places.) Confidence interval to b-2. Can we conclude that the packaging machine is operating improperly? Yes, since the confidence interval contains the target filling weight of 2.65. Yes, since the confidence interval does not contain the target filling weight of 2.65. No, since the confidence interval contains the target filling weight of 2.65. No, since the confidence interval does not contain the target filling weight of 2.65. c. How large a sample must we take if we want the margin of error to be at most 0.01 pound with 99% confidence? (Round intermediate calculations to 4 decimal places. Round "z-value" to 3 decimal places and round up your final answer to the next whole number.) Sample size

Explanation / Answer

a-1)The parameter of interest is the average filling weight of all cereal packages

a-2)point estimate =2.65

here std error =std deviation/(n)1/2 =0.06/(49)1/2 =0.0086

here for 99% CI ; z =2.576

hence margin of error E =z*std error =0.02

b-1) 99% confidence interval =point estimate -/+ z*std error = 2.63 to 2.67

b-2)No, since the confidence interval contains the target filling weight of 2.65

c)

margin of error E =0.01

for 99% CI ; z=2.5758

hence sample size n=(z*std deviation/E)2 =~ 239

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