In a random sample of eight cell phones, the mean full retail price was $559.00 and...
In a random sample of eight cell phones, the mean full retail price was $460.00 and the standard deviation was $173.00. 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 u. Interpret the results. Identify the margin of error. (Round to one decimal place as needed.) Construct a 95% confidence interval for the population mean. (Round to one decimal place as needed.) Interpret...
1) In a random sample of eight cell phones, the mean full retail price was $526.50 and the standard deviation was $184.00. Construct a 95% confidence interval for the true population price. Assume the sample was random and that the population of cell phone prices is normally distributed. Round your final answer to the nearest cent (2 numbers after the decimal)
In a random sample of eleven cell phones, the mean full retail price was $460.50 and the standard deviation was $190.00. Assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 90% confidence interval for the population mean mu. Interpret the results. Identify the margin of error.
In a random sample of eleven cell phones, the mean full retail price was $515.50 and the standard deviation was $212.00. Assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 90% confidence interval for the population mean mu. Interpret the results.
In a random sample of 7 cell phones, the mean full retail price was $502.40 and the standard deviation was $190.00 Further research suggests that the population mean is $427.17. Does the t-value for the original sample fall between −t 0.99 and t 0.99? Assume that the population of full retail prices for cell phones is normally distributed.
In a random sample of ten people, the mean driving distance to work was 18.6 miles and the standard deviation was 6.5 miles. Assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 90% confidence interval for the population mean . Interpret the results. Identify the margin of error. (Round to one decimal place as needed.) Construct a 90% confidence interval for the population mean (Round to one decimal place as...
In a random sample of 29 people, the mean commute time to work was 323 minutes and the standard deviation was 72 minutes. Assume the population is normally distributed and use at distribution to construct a 99% confidence interval for the population mean . What is the margin of error of u? Interpret the results The confidence interval for the population MAAN (Round to ona decimal ACA 2 neded) D The margin of error of his (Round to ona decimal...
A random sample of forty-four 200-meter swims has a mean time of 3.62 minutes and the population standard deviation is 0.09 minutes. Construct a 90% confidence interval for the population mean time. Interpret the results. The 90% confidence interval is OD (Round to two decimal places as needed.) Interpret the results. Choose the correct answer below A. B. c. With 90% confidence, it can be said that the population mean time is not between the endpoints of the given confidence...
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence ntervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals A random sample of 45 home theater systems has a mean price of $138.00. Assume the population standard deviation is 516.40. Construct a 90% confidence interval for the population mean. The 90% confidence interval...
In a random sample of four mobile devices, the mean repair cost was $60.00 and the standard deviation was $14.00. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a 90% confidence interval for the population mean. Interpret the results. The 90% confidence interval for the population mean is (DO (Round to two decimal places as needed.) The margin of error is $ (Round to two decimal places as needed.) Interpret...