Question

A machine that is programmed to package 3.70 pounds of cereal is being tested for its accuracy. In a sample of 61 cereal boxes, the sample mean filling weight is calculated as 3.72 pounds. The population standard deviation is known to be 0.10 pound. (You may find it useful to reference the z table.) a. Compute the point estimate as well as the margin of error with 95% confidence for the average filing weight of all cereal packages. (Round final answers to 2 decimal places.) Point estimate Margin of error b. Calculate the 95% confidence interval. (Round final answers to 2 decimal places.) Confidence interval to c. How large a sample must we take if we want the margin of error to be at most 0.02 pound with 95% confidence? (Round up final answer to nearest whole number.) Sample size

          A machine that is programmed to package 3.70 pounds of cereal is being tested for its accuracy. In a sample of 61 cereal boxes, the sample mean filling weight is calculated as 3.72 pounds. The population standard deviation is known to be 0.10 pound. (You may find it useful to reference the z table.)
a. Compute the point estimate as well as the margin of error with 95% confidence for the average filing weight of all cereal packages. (Round final answers to 2 decimal places.)
Point estimate
Margin of error
b. Calculate the 95% confidence interval. (Round final answers to 2 decimal places.)
Confidence interval to
c. How large a sample must we take if we want the margin of error to be at most 0.02 pound with 95% confidence? (Round up final answer to nearest whole number.)
Sample size

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A machine that is programmed to package 3.70 pounds of cereal is being tested for its accuracy. In a sample of 61 cereal boxes, the sample mean filling weight is calculated as 3.72 pounds. The population standard deviation is known to be 0.10 pound. (You may find it useful to reference the z table.)
a. Compute the point estimate as well as the margin of error with 95% confidence for the average filing weight of all cereal packages. (Round final answers to 2 decimal places.)
Point estimate
Margin of error
b. Calculate the 95% confidence interval. (Round final answers to 2 decimal places.)
Confidence interval to
c. How large a sample must we take if we want the margin of error to be at most 0.02 pound with 95% confidence? (Round up final answer to nearest whole number.)
Sample size

Added by Susan W.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert David Nguyen

08/23/2022

Step 1

Point estimate = sample mean = 3.72 pounds Margin of error = z * (population standard deviation / sqrt(sample size)), where z is the z-score for 95% confidence level, which is 1.96. Margin of error = 1.96 * (0.10 / sqrt(61)) Margin of error ≈ 0.025 Show more…

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A machine that is programmed to package 3.70 pounds of cereal is being tested for its accuracy. In a sample of 61 cereal boxes, the sample mean filling weight is calculated as 3.72 pounds. The population standard deviation is known to be 0.10 pound. (You may find it useful to reference the z_() table.) a. Compute the point estimate as well as the margin of error with 95% confidence for the average filing weight of all cereal packages. (Round final answers to 2 decimal places.) able[[,],[Point estimate,],[Margin of error,]] b. Calculate the 95% confidence interval. (Round final answers to 2 decimal places.) Confidence interval to c. How large a sample must we take if we want the margin of error to be at most 0.02 pound with 95% confidence? (Round up final answer to nearest whole number.) Sample size A machine that is programmed to package 3.70 pounds of cereal is being tested for its accuracy. In a sample of 61 cereal boxes, the it useful to reference the z table.) a. Compute the point estimate as well as the margin of error with 95% confidence for the average filing weight of all cereal packages (Round final answers to 2 decimal places.) Point estimate Margin of error b.Calculate the 95% confidence interval.(Round final answers to 2 decimal places.) Confidence interval to c. How large a sample must we take if we want the margin of error to be at most O.02 pound with 95% confidence? Round up final answer to nearest whole number.) Sample size

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Transcript

00:01 Hi, i'm david and i'm here to have your answer your question.

00:03 Now let me bring up your question here.

00:06 Let me make it bigger.

00:09 In the question here, we are going to discuss about the confidence interval for the mean.

00:16 And in this question here, we have the machine that is programmed to package 6 .20 pounds on the cereal.

00:26 And we have the symbol size and it goes to the 36.

00:30 And we have the symbol mean xpar equal to the 6 .29 and we have the population standard division equal to the 0 .13.

00:41 Now in the question a1, identify the relevant parameter of the interest for this quantitative data.

00:48 The parameter of the interest, it will be the mean weight of the serial package.

00:55 So therefore we have here we have the two options.

01:00 The the first one will be the proportion that will be wrong because we are interested in the average.

01:06 Therefore the right one will be the second option.

01:10 In the question a2, we need to compute the point estimate as well as the merchant of the error.

01:19 Now let me remind to you that the confidence interval for the mean, it will be the point estimate x plus or minus the merchant of the error, where the merchant of the error equal to the z and 4 over 2.

01:34 Time with the sigma overscore rather end.

01:38 And here we will have the point estimate it will equal to the xpore and equal to the 6 .29.

01:52 Now this one will be at the 99 % confidence.

01:57 Then we can find the z unfar over 2...

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