Question

A researcher records the repair cost for 20 randomly selected stereos. A sample mean of $57.17...

A researcher records the repair cost for 20 randomly selected stereos. A sample mean of $57.17 and standard deviation of $19.26 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the stereos. Assume the population is approximately normal.

Step 2 of 2 :

Construct the 90%90% confidence interval. Round your answer to two decimal places.

Homework Answers

Answer #1

TOPIC:Confidence interval for the population mean.

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 researcher records the repair cost for 8 randomly selected washers. A sample mean of $60.46...
A researcher records the repair cost for 8 randomly selected washers. A sample mean of $60.46 and standard deviation of $18.36 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the washers. Assume the population is approximately normal. Step 2 of 2: Construct the 90% confidence interval. Round your answer to two decimal places.
A consumer affairs investigator records the repair cost for 44 randomly selected TVs. A sample mean...
A consumer affairs investigator records the repair cost for 44 randomly selected TVs. A sample mean of $54.41 and standard deviation of $28.89 are subsequently computed. Determine the 90% 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. Step 2 of 2: Construct the 90% confidence interval. Round your...
A manager records the repair cost for 14 randomly selected dryers. A sample mean of $88.34...
A manager records the repair cost for 14 randomly selected dryers. A sample mean of $88.34 and standard deviation of $19.22 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the dryers. 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.
For the next 2 questions, please read the description below. A student records the repair costs...
For the next 2 questions, please read the description below. A student records the repair costs for 30 randomly selected computers from a local repair shop where he works. A sample mean of $216.53 is subsequently computed. Assume that the population standard deviation is $15.86, and the students is constructing a confidence interval. What is the error of estimate for a 98% confidence interval? What is the lower bound for a 96% confidence interval?
In a random sample of 35 ?refrigerators, the mean repair cost was ?$117.00 and the population...
In a random sample of 35 ?refrigerators, the mean repair cost was ?$117.00 and the population standard deviation is ?$16.30 Construct a 90?% confidence interval for the population mean repair cost. Interpret the results.
3. A sample of 30 randomly selected TV sets has a repair cost mean of $...
3. A sample of 30 randomly selected TV sets has a repair cost mean of $ 356.25, with a standard deviation of $ $19.75. Find the margin of error E for a 95% confidence interval.
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 35 refrigerators, the mean repair cost was $29.00  and the population standard...
In a random sample of 35 refrigerators, the mean repair cost was $29.00  and the population standard deviation is $15.90 . Construct a 95% confidence interval for the population mean repair cost. Interpret the results. 95%  confidence interval is________ ?   (Round to two decimal places as needed.) Interpret your results. Choose the correct answer below. A. With 95 % confidence, it can be said that the confidence interval contains the sample mean repair cost. B. With 95 % confidence, it can be...
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...
Fifty-nine items are randomly selected from a population of 650 items. The sample mean is 37,...
Fifty-nine items are randomly selected from a population of 650 items. The sample mean is 37, and the sample standard deviation 2. Develop a 90% confidence interval for the population mean. (Round the final answers to 2 decimal places.)                  The confidence interval is between       and