Question

For 30 randomly selected Rolling Stones concerts, the mean gross earnings is 2.46 million dollars. Assuming...

For 30 randomly selected Rolling Stones concerts, the mean gross earnings is 2.46 million dollars. Assuming a population standard deviation gross earnings of 0.54 million dollars, obtain a 99% confidence interval for the mean gross earnings of all Rolling Stones concerts (in millions).

a) confidence interval ( ______ , ______ )
Math   Statistics-And-Probability   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
For 30 randomly selected Rolling Stones concerts, the mean gross earnings is 2.55 million dollars. Assuming...
For 30 randomly selected Rolling Stones concerts, the mean gross earnings is 2.55 million dollars. Assuming a population standard deviation gross earnings of 0.55 million dollars, obtain a 99% confidence interval for the mean gross earnings of all Rolling Stones concerts (in millions). Confidence interval: ( equation editor , equation editor ). b) Which of the following is the correct interpretation for your answer in part (a)? A. There is a 99% chance that the mean gross earnings of all...
A group of 35 randomly selected towns has a mean debt of 1.2 million dollars. Assume...
A group of 35 randomly selected towns has a mean debt of 1.2 million dollars. Assume the population standard deviation is 1.68 million dollars. Construct a 95% confidence interval for the population mean. (0.6434, 1.7566) (0.4562, 1.5135) (1.0094, 1.9940) (0.5779, 2.6451)
An online used car company sells second-hand cars. For 30 randomly selected transactions, the mean price...
An online used car company sells second-hand cars. For 30 randomly selected transactions, the mean price is 3000 dollars. Part a) Assuming a population standard deviation transaction prices of 150 dollars, obtain a 99% confidence interval for the mean price of all transactions. Please carry at least three decimal places in intermediate steps. Give your final answer to the nearest two decimal places, and use at least 3 decimal places in your critical value.
The thicknesses of 30 randomly selected linoleum tiles were found to have a variance of 2.46....
The thicknesses of 30 randomly selected linoleum tiles were found to have a variance of 2.46. Construct the 98% confidence interval for the population variance of the thicknesses of all linoleum tiles in this factory. Round your answers to two decimal places.
1. A statistician is interested in the gross earnings of several of her favorite bands. She...
1. A statistician is interested in the gross earnings of several of her favorite bands. She took a random sample of 30 of the Rolling Stones’ North American concerts, and found that the gross earnings averaged $2.27 million with a standard deviation of $0.5 million. One source suggests that the average gross earnings per concert for every stadium performance in North America is $2.11 million. Do the Rolling Stones earn more on average? Test at a 5% level of significance...
The annual earnings​ (in dollars) of 35 randomly selected microbiologists are shown in the data table....
The annual earnings​ (in dollars) of 35 randomly selected microbiologists are shown in the data table. Use the data to​ (a) find the sample​ mean, (b) find the sample standard​ deviation, and​ (c) construct a​ 98% confidence interval for the population mean. 100,120 78,877 47,291 91,583 75,121 80,645 73,932 98,863 84,015 61,972 77,514 44,293 81,500 80,342 82,913 67,480 86,795 93,267 63,760 51,858 60,770 63,913 73,936 67,943 57,731 75,142 56,547 94,344 55,132 50,889 46,904 65,700 77,764 60,236 89,045
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?
An online used car company sells second-hand cars. For 30 randomly selected transactions, the mean price...
An online used car company sells second-hand cars. For 30 randomly selected transactions, the mean price is 2000 dollars. Part a) Assuming a population standard deviation transaction prices of 190 dollars, obtain a 99% confidence interval for the mean price of all transactions. Please carry at least three decimal places in intermediate steps. Give your final answer to the nearest two decimal places. Confidence interval: (. , ) For each situation, state the null and alternative hypotheses: (Type "mu" for...
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...
Randomly selected 30 student cars have ages with a mean of 7 years and a standard...
Randomly selected 30 student cars have ages with a mean of 7 years and a standard deviation of 3.6 years, while randomly selected 23 faculty cars have ages with a mean of 5.9 years and a standard deviation of 3.5 years. 1. Use a 0.01 significance level to test the claim that student cars are older than faculty cars. (a) The test statistic is (b) The critical value is (c) Is there sufficient evidence to support the claim that student...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT