n a random sample of six mobile​ devices, the mean repair cost was $65.00 and the standard deviation was $11.50. 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

A research team surveyed 251 adult Ohioans and found the mean monthly grocery cost for a single person is $248.33 with a standard deviation of $21.43. b. SUPPOSE the 90% confidence interval of the mean is $240 to $256. Interpret this confidence interval.   a. One can be 90% confident the true mean cost of groceries per month for a single person (in America) is between $240 and $256.   b. One can be 90% confident the sample mean cost of groceries per month for a single person (in Ohio) is between $240 and $256.   c. One can be 90% confident the true mean cost of groceries per month for a single person (in Ohio) is between $240 and $256.   d. 90% of adult Ohioans spend between $240 and $256 on their groceries every month.

In a random sample of eight people, the mean driving distance to work was 24 2 miles and the standard deviation was 7.5 miles. Assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 99% confidence interval for the population mean p. Interpret the results. Identify the margin of error. (Round to one decimal place as needed.)

An irate patient complained that the cost of a doctor's visit was too high. He randomly surveyed 20 other patients and found that the mean amount of money they spent on each doctor's visit was $48.60. The standard deviation of the sample is $3.53. a. b. Find the best point estimate of the population mean. Construct and interpret the 85% confidence interval of the mean amount of money spent on a doctor's visit.