43.) Americans have become increasingly concerned about the rising cost of Medicare. In 1990, the average annual Medicare spending per enrollee was $3267; in 2003, the average annual Medicare spending per enrollee was $6883 (Money, Fall 2003). Suppose you hired a consulting firm to take a sample of fifty 2003 Medicare enrollees to further investigate the nature of expenditures. Assume the population standard deviation for 2003 was $2,000 A. Calculate the standard error of the mean amount of Medicare spending for a sample of fifty 2003 enrollees (to 2 decimals). B. What is the probability the sample mean will be within +/- $300 of the population mean (to 4 decimals)? C. What is the probability the sample mean will be greater than $7500 (to 4 decimals)?

A) The standard error of the mean amount of Medicare spending for a sample of fifty 2003 enrollees is given by,

S.E = = 2,000/ = 282.8427= $282.84

B) Let be the sample mean. Then is normally distributed with mean $6883 and standard deviation = 2,000/ = $282.84Therefore Z = ( - 6883)/282.84 follows a Standard normal distribution. Now, the probability the sample mean will be within +/- $300 of the population mean is given by,

P[6583 < < 7183] = P[(6583- 6883)/282.84 < ( - 6883)/282.84 < (7183- 6883)/282.84]

= P[ -1.067 < Z < 1.067]= P[ Z < 1.067] – P[Z < -1.067]= 0.8556 – 0.1444= 0.7112

C) Now the probability the sample mean will be greater than $7500 is given by,P[ > 7500]

= P ( - 6883)/282.84 > (7500- 6883)/282.84]= P[ Z > 2.1814]= 1- P[Z < 2.1814]= 1 – 0.9854= 0.0146

45) The mean television viewing time for Americans is 15 hours per week (Money, November 2003). Suppose a sample of 40 Americans is taken to further investigate viewing habits. Assume the population standard deviation for weekly viewing time is = 3 hours. A. What is the probability the sample mean will be within 1 hour of the population mean (to 4 decimals)? B. What is the probability the sample mean will be within 45 minutes of the population mean (to 4 decimals)?

A) Let be the sample mean. Then is normally distributed with mean 15 hours and standard deviation = = 0.4743.Therefore Z = ( -15)/0.4743 follows a Standard normal distribution. Now the probability the sample mean will be within 1 hour of the population mean is given by,

P[14 < < 16] = P[(14 -15)/0.4743 < ( -15)/0.4743 < (16-15)/0.4743]= P[ -2.1082 < Z < 2.1082]= P[ Z < 2.1082] – P[Z < -2.1082]= 0.9825 – 0.0175= 0.9650

B) Now the probability the sample mean will be within 45 minutes of the population mean is given by,

P[14.25 < < 15.75] = P[(14.25 -15)/0.4743 < ( -15)/0.4743 < (15.75-15)/0.4743]= P[ -1.5811 < Z < 1.5811]= P[ Z < 1.5811] – P[Z < -1.5811]= 0.9431 – 0.0569= 0.8862