As part of an annual review of its accounts, a discount brokerage selects a random sample...
As part of an annual review of its accounts, a discount brokerage selects a random sample of 25 customers. Their accounts are reviewed for total account valuation, which showed a mean of $36,300, with a sample standard deviation of $8,950. (Use t Distribution Table.) What is a 90% confidence interval for the mean account valuation of the population of customers? (Round your answers to the nearest dollar amount.)
As part of an annual review of its accounts, a discount brokerage selects a random sample of 29 customers. Their accounts are reviewed for total account valuation, which showed a mean of $31,500, with a sample standard deviation of $8,650. (Use t Distribution Table.) What is a 95% confidence interval for the mean account valuation of the population of customers? (Round your answers to the nearest dollar amount.)
As a part of its annual review of accounts, a brokerage firm selects a random sample of 15 customers. The average account value for these customers was $32,000 with a standard deviation of $8,200. Determine the 95% confidence interval for the account value.
please include sketch, thanks! A discount brokerage selected a random sample of 64 customers and reviewed the value of their accounts. The mean was $32,000 with a population standard deviation of $8,200. What is a 90% confidence interval for the mean account value of the population of customers? Answers Point Estimate Standard Error Margin of Error Alpha Critical Value Confidence Interval
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....
A random sample of 13 customers in a Kentucky fast food restaurant revealed an average bill of OMR 18.75 per person. The population standard deviation is OMR5.4. Estimate 98% confidence interval for the mean bill of all customers. A random sample of ' n' customers in a Kentucky fast food restaurant revealed an average bill of OMR & per person. The population standard deviation is OMR 6. Estimate 98% confidence interval for the mean bill of all customers. n X...
A random sample of 28 customers at a gas station shows an average gas purchase of 8.9 gallons with a standard deviation of 3.2 gallons. Find the 98% confidence interval estimating the population mean number of gallons purchased at this station. Interpret this result.
A random sample of 144 checking accounts at a bank showed an average daily balance of $295. The standard deviation of the population is known to be $72. Please answer the following questions: (a) Find the standard error of the mean. (b) Give a point estimate of the population mean. (c) Construct a 95% confidence interval estimates for the mean.
a-c please 2. A random sample of 144 checking accounts at a bank showed an average daily balance of $295. The standard deviation of the population is known to be $72. Please answer the following questions: (a) Find the standard error of the mean. (b) Give a point estimate of the population mean. (c) Construct a 95% confidence interval estimates for the mean.
In a random sample of 8 people, the mean commute time to work was 33.5 minutes and the standard deviation was 7.4 minutes. A 98% confidence interval using thet-distribution was calculated to be (25.7,41.3). After researching commute times to work, it was found that the population standard deviation is 8.6 minutes. Find the margin of error and construct a 98% confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare the...