Question

The mean and standard deviation of a random sample of n measurements are equal to 33.9...

The mean and standard deviation of a random sample of n measurements are equal to 33.9 and 3.3, respectively.

a. Find the lower limit, the upper limit and the margin of error for a 95% confidence interval for m if n = 100.

b. Find the lower limit, the upper limit and the margin of error for  a 95% confidence interval for m if n = 400.

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
The mean and standard deviation of a random sample of n measurements are equal to 34.1...
The mean and standard deviation of a random sample of n measurements are equal to 34.1 and 3.6, respectively. a. Find a 90% confidence interval for μ if n=144. b. Find a 90% confidence interval for μ if n=576. c. Find the widths of the confidence intervals found in parts a and b. What is the effect on the width of a confidence interval of quadrupling the sample size while holding the confidence coefficient​ fixed?
A random sample of 49 measurements from a population with population standard deviation σ1 = 5...
A random sample of 49 measurements from a population with population standard deviation σ1 = 5 had a sample mean of x1 = 9. An independent random sample of 64 measurements from a second population with population standard deviation σ2 = 6 had a sample mean of x2 = 12. Test the claim that the population means are different. Use level of significance 0.01. (a) Compute the corresponding sample distribution value. (Test the difference μ1 − μ2. Round your answer...
A simple random sample with n=50provided a sample mean of 23.5 and a sample standard deviation...
A simple random sample with n=50provided a sample mean of 23.5 and a sample standard deviation of 4.3 a. Develop a 90% confidence interval for the population mean (to 1 decimal).   ,   b. Develop a 95% confidence interval for the population mean (to 1 decimal).   ,   c. Develop a 99% confidence interval for the population mean (to 1 decimal).   ,   d. What happens to the margin of error and the confidence interval as the confidence level is increased?
1)Given a sample mean is82, the sample size is 100and the population standard deviation is 20....
1)Given a sample mean is82, the sample size is 100and the population standard deviation is 20. Calculate the margin of error to 2 decimalsfor a 90% confidence level. 2)Given a sample mean is 82, the sample size is 100 and the population standard deviation is 20. Calculate the confidence interval for 90% confidence level. What is the lower limit value to 2 decimals? 3)Given a sample mean is 82, the sample size is 100 and the population standard deviation is...
A sample of n = 16 is to be taken from a distribution that can reasonably...
A sample of n = 16 is to be taken from a distribution that can reasonably be assumed to be Normal with a standard deviation σ of 100. The sample mean comes out to be 110. 1. The standard error of the mean, that is, the standard deviation of the sample mean, is σx¯ = σ/√ n. What is its numerical value? 2. The 97.5 percentile, 1.96, of the standard Normal distribution is used for a 95% confi- dence interval....
Consider a population having a standard deviation equal to 9.94. We wish to estimate the mean...
Consider a population having a standard deviation equal to 9.94. We wish to estimate the mean of this population. (a) How large a random sample is needed to construct a 95% confidence interval for the mean of this population with a margin of error equal to 1? (Round your answer to the next whole number.) The random sample is             units. (b) Suppose that we now take a random sample of the size we have determined in part a. If we...
A simple random sample with n=50 provided a sample mean of 22.5 and a sample standard...
A simple random sample with n=50 provided a sample mean of 22.5 and a sample standard deviation of 4.2 . a. Develop a 90% confidence interval for the population mean (to 1 decimal). ( , ) b. Develop a 95% confidence interval for the population mean (to 1 decimal). ( , ) c. Develop a 99% confidence interval for the population mean (to 1 decimal). ( , ) d. What happens to the margin of error and the confidence interval...
A simple random sample with n=56 provided a sample mean of 23.5 and a sample standard...
A simple random sample with n=56 provided a sample mean of 23.5 and a sample standard deviation of 4.5. a. Develop a 90% confidence interval for the population mean (to 1 decimal). (_____, ______) b. Develop a 95% confidence interval for the population mean (to 1 decimal). (_____, ______) c. Develop a 99% confidence interval for the population mean (to 1 decimal). (_____, ______) d. What happens to the margin of error and the confidence interval as the confidence level...
A simple random sample with N=52 provided a sample mean of 21.0 and a sample standard...
A simple random sample with N=52 provided a sample mean of 21.0 and a sample standard deviation of 4.4. a. Develop a 90% confidence interval for the population mean (to 1 decimal). (________, _________) b. Develop a 95% confidence interval for the population mean (to 1 decimal). (________, _________) c. Develop a 99% confidence interval for the population mean (to 1 decimal). (________, _________) d. What happens to the margin of error and the confidence interval as the confidence level...
In a random sample of 17 senior-level chemical engineers, the mean annual earnings was 123950 and...
In a random sample of 17 senior-level chemical engineers, the mean annual earnings was 123950 and the standard deviation was 34940. Assume the annual earnings are normally distributed and construct a 95% confidence interval for the population mean annual earnings for senior-level chemical engineers. 1. The critical value: 2. The standard error of the sample mean: 3. The margin of error: 4. The lower limit of the interval: 5. The upper limit of the interval: