Question

Use the given confidence level and sample data to find a confidence interval for the population...

Use the given confidence level and sample data to find a confidence interval for the population standard deviation . Assume that a simple random sample has been selected from a population that has a normal distribution. Salaries of college who took a course in college

​98% ​confidence n​=81 x^-= ​$65,700​   s=​$17,931

the table of​ Chi-Square critical values.

$ < standard deviation < $

Homework Answers

Answer #1

For n - 1 = 80 degrees of freedom, we have from chi square distribution tables here:

Also, we have here:

, Therefore,

Therefore the confidence interval for standard deviation here is computed as:

This is the required 98% confidence interval for population standard deviation here.

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
Use the given confidence level and sample data to find a confidence interval for the population...
Use the given confidence level and sample data to find a confidence interval for the population standard deviation sigma . Assume that a simple random sample has been selected from a population that has a normal distribution. Salaries of college professors who took a philosophy course in college 95 ​% ​confidence; n = 91​, x overbar = $65,200​, s = ​$16,815
Use the given confidence level and sample data to find a confidence interval for the population...
Use the given confidence level and sample data to find a confidence interval for the population standard deviation sigma. Assume that a simple random sample has been selected from a population that has a normal distribution. Salaries of college graduates who took a statistics course in college 80​% ​confidence; nequals61​, x overbarequals​$56 comma 600​, sequals​$16 comma 432
Use the given confidence level and sample data to find a confidence interval for the population...
Use the given confidence level and sample data to find a confidence interval for the population standard deviation. Assume that the population has a normal distribution. College students' annual earnings: 98% confidence; n = 9, sample mean = $3211, s = $897 Please show work using Excel
Use the given information to find the number of degrees of​ freedom, the critical values chi...
Use the given information to find the number of degrees of​ freedom, the critical values chi Subscript Upper L Superscript 2 and chi Subscript Upper R Superscript 2​, and the confidence interval estimate of sigma. It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. Nicotine in menthol cigarettes 98​% ​confidence; nequals21​, sequals0.28 mg. LOADING... Click the icon to view the table of​ Chi-Square critical values. dfequals nothing ​(Type a...
Use the given degree of confidence and sample data to construct a confidence interval for the...
Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ. Assume that the population has a normal distribution. Thirty randomly selected students took the calculus final. If the sample mean was 90 and the standard deviation s = 6.7, construct a 99% confidence interval for the mean score of all students
Find the following critical t-scores used in an 88% confidence interval for a population mean when...
Find the following critical t-scores used in an 88% confidence interval for a population mean when the population’s standard deviation (σσ) is unknown. Give each answer to at least three decimal places. The t-critical value for an 88% confidence interval with sample size 38 is: The t-critical value for an 88% confidence interval with sample size 48 is: The t-critical value for an 88% confidence interval with sample size 64 is: The t-critical value for an 88% confidence interval with...
Use the given information to find the number of degrees of​ freedom, the critical values chi...
Use the given information to find the number of degrees of​ freedom, the critical values chi Subscript Upper L Superscript 2χ2L and chi Subscript Upper R Superscript 2χ2R​, and the confidence interval estimate of sigmaσ. It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. Platelet Counts of Women 98% ​confidence; nequals=27​, sequals=65.7
Use the given information to find the number of degrees of​ freedom, the critical values chi...
Use the given information to find the number of degrees of​ freedom, the critical values chi Subscript Upper L Superscript 2 and chi Subscript Upper R Superscript 2​, and the confidence interval estimate of sigma. It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. Nicotine in menthol cigarettes 99​% ​confidence; n=23​, s=0.27 mg. Use the given information to find the number of degrees of​ freedom, the critical values chi...
Use the confidenve level and sample data to find a confidence interval for estimating the population...
Use the confidenve level and sample data to find a confidence interval for estimating the population mean. A special research company randomly selected a simple random sample of 75 birth weights in US. The mean has of 3433 g with population standard deviation of 495 g of all birth weights in US. Construct a 90% confidence interval to estimate the population mean of birth weights Initial Data: E(margin of error result value) = Cl(population mean confidence interval result value)=
use the given information to find the number of degrees of freedom the critical values x2l...
use the given information to find the number of degrees of freedom the critical values x2l and x2r and the confidence interval estimate of the standard deviation it is reasonable to assume that a simple random sample has been selected from a population with a normal distribution platelet counts of women 98% confidence in equals 28 s equals 65.5 what is the degrees of freedom...X2L...X2R..CI estimate of the standard deviation?
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT