Question

The weights of 6 randomly selected mattresses were found to have a variance of 1.48. Construct...

The weights of 6 randomly selected mattresses were found to have a variance of 1.48. Construct the 99% confidence interval for the population variance of the weights of all mattresses in this factory. Round your answers to two decimal places.

(lower and upper endpoint)

0 0
Add a comment Improve this question Transcribed image text
Answer #1

GIVEN:

Sample size of mattresses

Sample variance

FORMULA USED:

The formula for confidence interval for the population variance is,

CALCULATION:

SIGNIFICANCE LEVEL:

Significance level = 1 - Confidence level

= 1 - 0.99

= 0.01

CRITICAL VALUE:

At degrees of freedom,

CONFIDENCE INTERVAL:

The 99% confidence interval for the population variance of the weights of all mattresses in this factory is,

Thus the 99% confidence interval for the population variance of the weights of all mattresses in this factory is .

Add a comment
Know the answer?
Add Answer to:
The weights of 6 randomly selected mattresses were found to have a variance of 1.48. Construct...
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. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • The weights of 89randomly selected mattresses were found to have a standard deviation of 3.78 Construct...

    The weights of 89randomly selected mattresses were found to have a standard deviation of 3.78 Construct the 90% confidence interval for the population standard deviation of the weights of all mattresses in this factory. Round your answers to two decimal places.

  • The thicknesses of 55 randomly selected ceramic tiles were found to have a variance of 3.18...

    The thicknesses of 55 randomly selected ceramic tiles were found to have a variance of 3.18 Construct the 98% confidence interval for the population variance of the thicknesses of all ceramic tiles in this factory. Round your answers to two decimal places.

  • Step 2. Suppose a sample of 1134 tenth graders is drawn. Of the students sampled, 930...

    Step 2. Suppose a sample of 1134 tenth graders is drawn. Of the students sampled, 930 read above the eighth grade level. Using the data, construct the 99% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level. Round your answers to three decimal places. Answer: Lower endpoint: Upper endpoint: The thicknesses of 81 randomly selected aluminum sheets were found to have a variance of 3.23. Construct the 98% confidence interval for the...

  • here are summary statistics for randomly selected weights of newborn girls: n=235, 2 of 6 (1...

    here are summary statistics for randomly selected weights of newborn girls: n=235, 2 of 6 (1 complete) HW Score: 16.67%, 1 of 6 Score: 0 of 1 pt 7.2.9-T Question Help Here are summary statistics for randomly selected weights of newborn girls: n=235, x = 27.6 hg, s = 7.6 hg. Construct a confidence interval estimate of the mean Use a 99% confidence level. Are these results very different from the confidence interval 25.9 g < 30.5 hg with only...

  • A supervisor records the repair cost for 17 randomly selected TVs. A sample mean of $57.20...

    A supervisor records the repair cost for 17 randomly selected TVs. A sample mean of $57.20 and standard deviation of $26.41 are subsequently computed. Determine the 80% confidence interval for the mean repair cost for the TVS. Assume the population is approximately normal. Step 1 of 2: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places. Answer/How to Enter) 2 Points Tables Keypad Keyboard Shortcuts A supervisor records the...

  • Here are summary statistics or randomly selected weights of newborn girls

    Here are summary statistics or randomly selected weights of newborn girls: n-186, x-265 hg.-68g Construct a confidence interval estimate of the mean, Use a 99% confidence level Are these results very different from the confidence interval 24.6 hg < μ 27.6 hg with only 19 sample values x = 26.1 hg and s=2.3 hg? What is the confidence interval for the population mean μ? (Round to one decimal place as needed) 

  • A survey of 25 randomly selected customers found the ages shown (in years). The mean is...

    A survey of 25 randomly selected customers found the ages shown (in years). The mean is 32.00 years and the standard deviation is 9.99 3 years. a Construct a 99% confidence interval for the mean age of all customers, assuming that the assumptions and conditions or the con interval have been met b) How large is the margin of error? c) How would the confidence interval change i you had assumed that the standard deviation was known to be 10.0...

  • The following data were randomly drawn from an approximately normal population. 37,38, 40, 44, 47,50 Based...

    The following data were randomly drawn from an approximately normal population. 37,38, 40, 44, 47,50 Based on these data, find a 99% confidence interval for the population standard deviation. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. What is the lower limit of the 99% confidence interval? What is the upper limit of the 99% confidence interval? x 5 ?

  • In a poll of 125 randomly selected U.S. adults, 67 said they favored a new proposition....

    In a poll of 125 randomly selected U.S. adults, 67 said they favored a new proposition. Based on this poll, compute a 99% confidence interval for the proportion of all U.S. adults in favor of the proposition (at the time of the poll). Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. What is the lower limit of the 99% confidence interval? What is the upper limit...

  • In a poll of 175 randomly selected U.S. adults, 92 said they favored a new proposition....

    In a poll of 175 randomly selected U.S. adults, 92 said they favored a new proposition. Based on this poll, compute a 99% confidence interval for the proportion of all U.S. adults in favor of the proposition (at the time of the poll). Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. (If necessary, consult a list of formulas.) What is the lower limit of the 99%...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT