In a random sample of four mobile devices, the mean repair cost was $70.00 and the standard deviation was $14.00. Assume the population is norm 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 u is ( ). (Round to two decimal places as needed.) The margin of error is $ (Round to two decimal places as needed.) Interpret the results. Choose the correct answer below. O A. With 90% confidence, it can be said that the population mean repair cost is between the bounds of the confidence interval, O B. It can be said that 90% of mobile devices have a repair cost between the bounds of the confidence interval. O C. With 90% confidence, it can be said that the repair cost is between the bounds of the confidence interval. O D. If a large sample of mobile devices are taken approximately 90% of them will have repair costs between the bounds of the confidence interval.

The distribution of study times for first-year students follow a Normal distribution with a standard deviation σ = 48 minutes. Suppose a sample of 344 students found that the mean study time for these students was x¯= 170 minutes per week. Use this to find a 94.38% confidence interval to estimate the mean study time for all students. We are 94.38% confident that the mean study time for first-year students is between the following ( answer) minutes and  ( answer) minutes. The answer needs to be the mean study time for first-year students is between what amount of minutes and the following minutes.

The standard deviation, S, for the annual medical cost of 24 households was found to equal $1689 find a 95% confidence interval for the standard deviation of the annual medical cost for all households. Assume medical costs are normally distributed.

Assume that the heights of women are normally distributed. A random sample of 20 women have a mean height of 62.5 inches and a standard deviation of 1.3 inches. Construct a 98% confidence interval for the population variance, .