Question

In estimating the mean monthly fuel expenditure, per household, a random sample of how many must...

In estimating the mean monthly fuel expenditure, per household, a random sample of how many must be selected to be 99% confident and a population s. d. of 1.65, and margin of error, 5%.

Homework Answers

Answer #1

Solution :

Given that,

Population standard deviation = = 1.65

Margin of error = E = 0.05

At 99% confidence level

= 1 - 99%  

= 1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z0.005 = 2.576

sample size = n = [Z/2* / E] 2

n = [ 2.576 * 1.65 / 0.05 ]2

n = 7226.36

Sample size = n = 7327

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 decision maker is interested in estimating the mean of a population based on a random...
A decision maker is interested in estimating the mean of a population based on a random sample. She wants the confidence level to be 90​% and the margin of error to be ±0.50 She does not know what the population standard deviation​ is, so she has selected the pilot sample below. Based on this pilot​ sample, how many more items must be sampled so that the decision maker can make the desired confidence interval​ estimate? 8.50 4.87 10.95 15.05 14.78...
A research study aims to estimate the mean annual household expenditure of all households in a...
A research study aims to estimate the mean annual household expenditure of all households in a metropolitan city. The study investigator plans to collect data on annual expenditure from a random sample of 100 households. By studying the sampling distribution of the mean annual expenditure, we will be able to ... (choose all that apply) a) compute the mean annual household expenditure of all households in the metropolitan city; b) obtain all possible random samples of households from the metropolitan...
A simple random sample of electronic components will be selected to test for the mean lifetime...
A simple random sample of electronic components will be selected to test for the mean lifetime in hours. Assume that component lifetimes are normally distributed with population standard deviation of 31 hours. How many components must be sampled so that a 99% confidence interval will have margin of error of 6 hours? Write only an integer as your answer.
5. Mean of 1500, standard deviation of 300. Estimating the average SAT score, limit the margin...
5. Mean of 1500, standard deviation of 300. Estimating the average SAT score, limit the margin of error to 95% confidence interval to 25 points, how many students to sample? 6. Population proportion is 43%, would like to be 95% confident that your estimate is within 4.5% of the true population proportion. How large of a sample is required? 7. P(z<1.34) (four decimal places) 8. Candidate only wants a 2.5% margin error at a 97.5% confidence level, what size of...
4) On a standard IQ test, σ = 15. How many random IQ scores must be...
4) On a standard IQ test, σ = 15. How many random IQ scores must be obtained if we want to find the true population mean, with an allowable error of 0.5 and we want 99% confidence in the results? 5) Given n = 25, s = 15, and α = 0.01, find the maximum error of estimate, E.
How much does household weekly income affect the household weekly expenditure on food? The following data...
How much does household weekly income affect the household weekly expenditure on food? The following data shows household weekly expenditure on food and the household weekly income (all in dollars). Use the data below to develop an estimated regression equation that could be used to predict food expenditure for a weekly income. Use Excel commands for your calculations. FOOD INCOME y x 91 292 148 479 107 428 146 766 243 1621 312 1661 243 1292 272 1683 349 1808...
Use the confidence level and sample data to find a confidence interval for estimating the population...
Use the confidence level and sample data to find a confidence interval for estimating the population mean. A local telephone company randomly selected a sample of 679 duration of calls. The a sample mean amount was 6.81 min with a population standard deviation of 5.1 min. Construct a confidence interval to estimate the population mean of pulse rate if a confidence level is 99% . Initial Data: E(margin of error result value) = CI(population mean confidence interval result value)=
2. Market researchers want to estimate the mean monthly expenditure on transport for households in Sydney....
2. Market researchers want to estimate the mean monthly expenditure on transport for households in Sydney. They randomly selected 19 householders and asked their expenditure on transport, in dollars, over the past month. Data from the selected households are given below. 181 266 339 292 527 150 484 488 282 571 536 153 583 434 135 187 497 144 520 Part A What is the best point estimate of the population mean, to two decimal places? Give your answer in...
A simple random sample of size n is drawn. The sample​ mean, x overbar​, is found...
A simple random sample of size n is drawn. The sample​ mean, x overbar​, is found to be 17.6​, and the sample standard​ deviation, s, is found to be 4.1. LOADING... Click the icon to view the table of areas under the​ t-distribution. ​(a) Construct a 95​% confidence interval about mu if the sample​ size, n, is 35. Lower​ bound: nothing​; Upper​ bound: nothing ​(Use ascending order. Round to two decimal places as​ needed.) ​(b) Construct a 95​% confidence interval...
A simple random sample of size n is drawn. The sample​ mean, x overbar​, is found...
A simple random sample of size n is drawn. The sample​ mean, x overbar​, is found to be 17.7​, and the sample standard​ deviation, s, is found to be 4.8. Click the icon to view the table of areas under the​ t-distribution. ​(a) Construct a 95​% confidence interval about mu if the sample​ size, n, is 35. Lower​ bound: nothing​; Upper​ bound: nothing ​(Use ascending order. Round to two decimal places as​ needed.) ​(b) Construct a 95​% confidence interval about...