Question

Under the original process, 50 components were manufactured with a net profit per component of $1,255...

Under the original process, 50 components were manufactured with a net profit per component of $1,255 with a standard deviation of $215. Under the new process, 30 components were manufactured with a net profit of $1,330 with a standard deviation of $238. Calculate the 95% confidence interval.

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A random sample of 380 electronic components manufactured by a certain process are tested, and 30...
A random sample of 380 electronic components manufactured by a certain process are tested, and 30 are found to be defective. A device will be manufactured in which two of the components will be connected in series. The components function independently, and the device will function only if both components function. Let q be the probability that a device functions. Find a 95% confidence interval for q. Round the answers to three decimal places.
A random sample of 400 electronic components manufactured by a certain process are tested, and 30...
A random sample of 400 electronic components manufactured by a certain process are tested, and 30 are found to be defective. The company ships the components in lots of 200. Lots containing more than 20 defective components may be returned. Find a 95% confidence interval for the proportion of lots that will be returned. Use the normal approximation to compute this proportion. Round the answers to four decimal places. The 95% confidence interval is (, ).? The first answer is...
n an accelerated failure test, components are operated under extreme conditions so that a substantial number...
n an accelerated failure test, components are operated under extreme conditions so that a substantial number will fail in a rather short time. In such a test involving two types of microchips, 580 chips manufactured by an existing process were tested, and 125 of them failed. Then, 780 chips manufactured by a new process were tested, and 130 of them failed. Find a 90% confidence interval for the difference between the proportions of failures for chips manufactured by the two...
A modification has been made to the production line which produces a certain electronic component. Past...
A modification has been made to the production line which produces a certain electronic component. Past evidence shows that lifetimes of the components have a population standard deviation of 700 hours. It is believed that the modification will not affect the population standard deviation, but may affect the mean lifetime. The previous mean lifetime was 3250 hours. A random sample of 50 components made with the modification is taken and the sample mean lifetime for these components is found to...
in a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with...
in a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with garlic in a process tablet form. cholesterol levels were measured before and after the treatment. the changes( before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 5.6 and a standard deviation of 18.1. Construct a 95% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. what does the confidence interval suggest about...
50 Football Players were observed for a race. The sample mean was 30 seconds and the...
50 Football Players were observed for a race. The sample mean was 30 seconds and the sample standard deviation was 2 seconds. Assuming that the population standard deviation is unknown, what is the 95% confidence interval for the population mean? A) (26, 34) B) (26.08, 33.92) C) (29.446, 30.554) D) (29.432, 30.568)
In order to test a new production method, 15 employees were selected randomly to try the...
In order to test a new production method, 15 employees were selected randomly to try the new method. The mean production rate for the sample was 80 parts per hour, and the standard deviation was calculated to be 10 parts per hour. CHAPTER 7. CONFIDENCE INTERVALS 120 a. Give a 95% confidence interval for the true mean production rate for the new method. b. Interpret the interval in part a. c. Suppose the confidence interval in part a is considered...
In a test of the effectiveness of garlic for lowering​ cholesterol, 50 subjects were treated with...
In a test of the effectiveness of garlic for lowering​ cholesterol, 50 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes ​(beforeminus​after) in their levels of LDL cholesterol​ (in mg/dL) have a mean of 3.2 and a standard deviation of 18.5. Construct a 95​% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness...
Moen has instituted a new manufacturing process for assembling a component. Using the old process, the...
Moen has instituted a new manufacturing process for assembling a component. Using the old process, the average assembly time was 4.25 minutes per component. After the new process was in place for 30 days, the quality control engineer took a random sample of fourteen components and noted the time it took to assemble each component. Based on the sample data, the average assembly time was 4.38 minutes, with a standard deviation of .16 minutes. Using a significance level of .02,...
In a test of the effectiveness of garlic for lowering cholesterol 50 subjects were treated to...
In a test of the effectiveness of garlic for lowering cholesterol 50 subjects were treated to with garlic in the process tablet form cholesterol levels were measured Before & After the treatment the changes Before & After and their levels of LDL cholesterol in MG/DL have a mean of 5.4 and a standard deviation of 18.3 construct a 99% confidence interval estimate of the mean net change in LDL cholesterol after the girl it treatment what does this confidence interval...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT