Question

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 needed) Interpret the results. Select the correct choice below and fill in the answer box to complete your choice (Type an integer or a decimal. Do not round.) O A It can be said that % of the population has a driving distance to work in miles) that is between the interval's endpoints. O B. With % confidence, it can be said that most driving distances to work in miles) in the population are between the interval's endpoints. O C. With % confidence, it can be said that the population mean driving distance to work in miles) is between the interval's endpoints. OD. % of all random samples of ten people from the population will have a mean driving distance to work in miles) that is between the interval's endpoints.

Answer 1

nalo, = 18.6, 5 = 6.5, C= 9004 Q = 1-0.90=0.1 1df-fi-| = lor 212= 0.05 tarla, dfa t0.05,9 = +11833 [osing t teble] man margin of error ME = ta 12, of x 5 / 5 = 1.833 % 65 = 3.767696 margin of error = (3.8 . 90 % Confidence interval for the population mean. = Ă + ME = 18.6 + 3.767696 = 14,8323, 92.3677 90.7.(.I ={14.8] , [22.4) with 90 %. Confidence, it Gn be said that the population mean driving distance to work (in miles) is blew the interval's end boints. XO