Answer:-
Given that:-
monthly income in $1000
4.0, 5.0, 7.0, 4.0 , 6.0, 6.0, 7.0, 9.0
mean
(in $1000)
(a) standard of mean
($1000)
Standard error of mean = 558.6 $
b)
Margin of error,
($1000)
c) 95% Confidence interval is given by
(in $1000)
in $
95% C.I for the mean of population is ($4,682 , $ 7,318)
4. The monthly incomes from a random sample of workers in a factory are shown below....
The monthly incomes from a random sample of workers in a factory are shown below. Monthly Income, in $1,000 4.0 5.0 7.0 5.0 6.0 6.0 10.0 8.0 9.0 (Please, avoid rounding intermediate steps and round your final solutions to at least 2 decimal places) Compute the 95% confidence interval for the mean monthly incomes of the workers. Provide the lower and upper bound of the interval below and give your answer in dollars. Are there any additional assumptions needed in...
The monthly incomes from a random sample of faculty at a university are shown below. Monthly Income ($1000s) 3.0 4.0 6.0 3.0 5.0 5.0 6.0 8.0 Compute a 90% confidence interval for the mean of the population. The population of all faculty incomes is known to be normally distributed. Give your answer in dollars.
Suppose a random sample of size 17 was taken from a normally distributed population, and the sample standard deviation was calculated to be s = 5.0. a) Calculate the margin of error for a 95% confidence interval for the population mean. Round your response to at least 3 decimal places. b) Calculate the margin of error for a 90% confidence interval for the population mean. Round your response to at least 3 decimal places.
6.2.19-T Question Help In a random sample of four microwave ovens, the mean repair cost was $85.00 and the standard deviation was $13.00. 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. The 99% confidence interval for the population mean μ is (DD (Round to two decimal places as needed.) 6.2.21-T Question Help In a random sample...
In a random sample of 8 people with advanced degrees in biology, the mean monthly income was $4744 and the standard deviation was $580. Assume the monthly incomes are normally distributed. Construct a 95% confidence interval for the population mean monthly income for people with advanced degrees in biology.
In a random sample of 11 people, the mean driving distance to work was 25.2 miles and the standard deviation was 7.3 miles. Assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 95% confidence interval for the population mean. Identify margin of error Construct a 95% confidence interval for the population mean (___,___)
In a random sample of 8 people, the mean commute time to work was 34.5 minutes and the standard deviation was 7.3 minutes. A 95% confidence interval using the t-distribution was calculated to be (28.4.40.6). After researching commute times to work, it was found that the population standard deviation is 9.4 minutes. Find the margin of error and construct a 95% confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare...
Suppose that a simple random sample of size ?=325 selected from a population has ?=147 successes. Calculate the margin of error for a 95% confidence interval for the proportion of successes for the population, ? . Compute the sample proportion, ?̂, standard error estimate, SE, critical value, ?, and the margin of error, ?. Use a ?- distribution table to determine the critical value. Give all of your answers to three decimal places except give the critical value, ?, to...
1. A random sample of 82 customers, who visited a department store, spent an average of $71 at this store. Suppose the standard deviation of expenditures at this store is O = $19. What is the e 98% confidence interval for the population mean? 2. A sample of 25 elements produced a mean of 123.4 and a standard deviation of 18.32 Assuming that the population has a normal distribution, what is the 90% confidence interval for the population mean? 3....
In a random sample of six mobile devices, the mean repair cost was $80.00 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