Question

A survey of 37 randomly selected iPhone owners showed that the purchase price has a mean...

A survey of 37 randomly selected iPhone owners showed that the purchase price has a mean of $670 with a sample standard deviation of $34. (Use z Distribution Table.)

  1. Compute the standard error of the sample mean. (Round your answer to the nearest whole number.)
  1. Compute the 99% confidence interval for the mean. (Use t Distribution Table.) (Round your answers to 3 decimal places.)

Homework Answers

Answer #1

Sample size, n = 37

mean = $670

sample standard deviation = s = $34

Compute the standard error of the sample mean.

Compute the 99% confidence interval for the mean.

At 99% confidence interval and df = n - 1 = 37 - 1 = 36

t_c = 2.7195

The 99% confidence interval is given by:

CI = ( 670 - 15.2, 670 +15.2)

CI = ( 654.8, 685.2 )

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 survey of 33 randomly selected iPhone owners showed that the purchase price has a mean...
A survey of 33 randomly selected iPhone owners showed that the purchase price has a mean of $630 with a sample standard deviation of $26. Compute the standard error of the sample mean Compute the 99% confidence interval for the mean. To be 99% confident, how large a sample is needed to estimate the population mean within $11?
A survey of 25 randomly selected customers had the mean 33.04 years and the standard deviation...
A survey of 25 randomly selected customers had the mean 33.04 years and the standard deviation 10.45 years. What is the standard error of the mean? How would the standard error change if the sample size had been             400 instead of 25? (Assume that the sample standard deviation didn't change.) Find the degrees of freedom for n=60. Find the t critical value for n=60 for 90% CI for mean. A survey of 25 randomly selected customers had the mean...
A survey of 20 randomly selected adult men showed that the mean time they spend per...
A survey of 20 randomly selected adult men showed that the mean time they spend per week watching sports on television is 9.34 hours with a standard deviation of 1.34 hours. Assuming that the time spent per week watching sports on television by all adult men is (approximately) normally distributed, construct a 90 % confidence interval for the population mean,   μ . Round your answers to two decimal places. Lower bound: Enter your answer; confidence interval, lower bound Upper bound:...
A survey of 100 randomly selected Michigan homeowners finds that they spend a mean of $67...
A survey of 100 randomly selected Michigan homeowners finds that they spend a mean of $67 per month on home maintenance. In an independent survey, 100 randomly selected Florida homeowners find that they spend a mean of $75 per month on home maintenance. Construct a 99% confidence interval for the true difference in the mean amounts of money spent per month on home maintenance for the two states. Assume that the population standard deviation is $14 per month for Michigan...
Confidence Interval Examples 1. A recent survey based on 100 TCOB undergraduates showed that 40% of...
Confidence Interval Examples 1. A recent survey based on 100 TCOB undergraduates showed that 40% of the respondents were using an Iphone. What is the 90% confidence interval for the proportion of Iphone users among TCOB undergraduates? (1) What is the corresponding Z-­‐value? (2) What is the sampling error for the proportion? (3) What is the confidence interval? 2. A sample study of 100 customers of Best Buy showed that the average dollar spending during the last three months was...
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 32.52 years and the standard deviation is 9.40 years. ​ 37 33 38 24 48 41 39 38 15 26 20 41 37 25 11 28 35 33 28 31 40 30 48 42 25 a) Construct a 99​% confidence interval for the mean age of all​ customers, assuming that the assumptions and conditions for the confidence interval have been met. (__,__) ​b) How...
A sample of 250 observations is selected from a normal population with a population standard deviation...
A sample of 250 observations is selected from a normal population with a population standard deviation of 23. The sample mean is 18. Determine the standard error of the mean. (Round your answer to 3 decimal places.) Determine the 99% confidence interval for the population mean. (Use z Distribution Table.) (Round your answers to 3 decimal places.)
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
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 31.04 years and the standard deviation is 9.00 years.​ a) Construct a 99​% confidence interval for the mean age of all​ customers, assuming that the assumptions and conditions for the confidence interval have been met. ​b) How large is the margin of​ error? ​c) How would the confidence interval change if you had assumed that the standard deviation was known to be 9.0 ​years?
A survey is being planned to determine the mean amount of time corporation executives watch television....
A survey is being planned to determine the mean amount of time corporation executives watch television. A pilot survey indicated that the mean time per week is 15 hours, with a standard deviation of 2.5 hours. It is desired to estimate the mean viewing time within 30 minutes. The 99% level of confidence is to be used. (Use z Distribution Table.) How many executives should be surveyed? (Round your z-score to 2 decimal places and round up your final answer...