Question

In a random sample of six mobile​ devices, the mean repair cost was $75.00 and the...

In a random sample of six mobile​ devices, the mean repair cost was $75.00 and the standard deviation was $13.0013.00. Assume the population is normally distributed and use a​ t-distribution to find the margin of error and construct a 95​% confidence interval for the population mean. Interpret the results.

The 95​% confidence interval for the population mean μ is

Homework Answers

Answer #1

Solution :

Given that,

Point estimate = sample mean = = $75.00

sample standard deviation = s = $13.00

sample size = n = 6

Degrees of freedom = df = n - 1 = 6 - 1 = 5

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,5 = 2.571

Margin of error = E = t/2,df * (s /n)

= 2.571 * (13.00 / 6)

Margin of error = E = 13.64

The 95% confidence interval estimate of the population mean is,

- E < < + E

75.00 - 13.64 < < 75.00 + 13.64

61.36 < < 88.64

The 95​% confidence interval for the population mean μ is (61.36 , 88.64)

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
In a random sample of six mobile devices, the mean repair cost was $60.00 and the...
In a random sample of six mobile devices, the mean repair cost was $60.00 and the standard deviation was $12.50. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a 99% confidence interval for the population mean. Interpret the results. The 99% confidence interval for the population μ is (_, _)
In a random sample of four mobile​ devices, the mean repair cost was ​$65.00 and the...
In a random sample of four mobile​ devices, the mean repair cost was ​$65.00 and the standard deviation was ​$13.50. Assume the population is normally distributed and use a​ t-distribution to find the margin of error and construct a 99​% confidence interval for the population mean. Interpret the results.
In a random sample of four mobile​ devices, the mean repair cost was ​$85.00 and the...
In a random sample of four mobile​ devices, the mean repair cost was ​$85.00 and the standard deviation was ​$12.00. Assume the population is normally distributed and use a​ t-distribution to find the margin of error and construct a 95​% confidence interval for the population mean. Interpret the results. The 95​% confidence interval for the population mean mu is ​( ​,​). ​(Round to two decimal places as​ needed.) The margin of error is ​$ nothing. ​(Round to two decimal places...
In a random sample of 13 microwave​ ovens, the mean repair cost was ​$70.00 and the...
In a random sample of 13 microwave​ ovens, the mean repair cost was ​$70.00 and the standard deviation was ​$15.40. Using the standard normal distribution with the appropriate calculations for a standard deviation that is​ known, assume the population is normally​ distributed, find the margin of error and construct a 98​% confidence interval for the population mean. A 98​% confidence interval using the​ t-distribution was (58.5,81.5). Find the margin of error of the population mean. Find the confidence interval of...
In a random sample of six ​people, the mean driving distance to work was 18.3 miles...
In a random sample of six ​people, the mean driving distance to work was 18.3 miles and the standard deviation was 4.3 miles. Assume the population is normally distributed and use the​ t-distribution to find the margin of error and construct a 90​% confidence interval for the population mean mu. Interpret the results. Identify the margin of error.
In a random sample of 29 ​people, the mean commute time to work was 30.1 minutes...
In a random sample of 29 ​people, the mean commute time to work was 30.1 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a​ t-distribution to construct a 80​% confidence interval for the population mean μ? What is the margin of error of μ​? Interpret the results.
In a random sample of 22 ​people, the mean commute time to work was 32.1 minutes...
In a random sample of 22 ​people, the mean commute time to work was 32.1 minutes and the standard deviation was 7.3 minutes. Assume the population is normally distributed and use a​ t-distribution to construct a 99​% confidence interval for the population mean μ. What is the margin of error of μ​? Interpret the results.
In a random sample of 28 ​people, the mean commute time to work was 33.6 minutes...
In a random sample of 28 ​people, the mean commute time to work was 33.6 minutes and the standard deviation was 7.1 minutes. Assume the population is normally distributed and use a​ t-distribution to construct a 99​% confidence interval for the population mean μ. What is the margin of error of μ​? Interpret the results.
1) Use the given confidence interval to find the margin of error and the sample mean....
1) Use the given confidence interval to find the margin of error and the sample mean. ​(12.8​,19.8​) 2) In a random sample of four microwave​ ovens, the mean repair cost was ​$60.00 and the standard deviation was ​$12.00. Assume the population is normally distributed and use a​ t-distribution to construct a 99​% confidence interval for the population mean ?. What is the margin of error of ? Interpret the results.
In a random sample of eight cell​ phones, the mean full retail price was ​$450.00 and...
In a random sample of eight cell​ phones, the mean full retail price was ​$450.00 and the standard deviation was ​$208.00. Assume the population is normally distributed and use the​ t-distribution to find the margin of error and construct a 90​% confidence interval for the population mean mu. Interpret the results. Identify the margin of error.