Question
Ch8 Question 45
A sample survey of 54 discount brokers showed that the mean price charged for a trade of 100 shares at exist50 per share was exist33.77. The survey is conducted annually. With the historical data available, assume a known population standard deviation of exist15. a. Using the sample data, what is the margin of error associated with a 95% confidence interval? b. Develop a 95% confidence interval for the mean price charged by discount brokers a trade of 100 shares at exist50 per share. A survey conducted by the American Automobile Association showed that a family of four spends an average of exist215.60 per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of exist252.45 per day and a sample standard deviation of exist74.50. a. Develop a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls. b. Based on the confidence interval from part (a), does it appear that the population mean amount spent per day by families visiting Niagara Falls differs from the mean reported by the American Automobile Association? Explain. The 92 million Americans of age 50 and over control 50 percent of all discretionary income. AARP estimates that the average annual expenditure on restaurants and carryout food was exist1873 for individuals in this age group. Suppose this estimate is based on a sample of 80 persons and that the sample standard deviation is exist550. a. At 95% confidence, what is the margin of error? b. What is the 95% confidence interval for the population mean amount spent restaurants and carryout food? c. What is your estimate of the total amount spent by Americans of age 50 and over on restaurants and carryout food? d. If the amount spent on restaurants and carryout food is skewed to the right, would you expect the median amount spent to be greater or less than exist1873?
Explanation / Answer
Question 45)
Part a)
Here the sample size is large, so we use z distribution to find the confidence interval.
At 5% level of significance the critical z is given by 1.96
X bar = 252.45
s = 74.50
n = 64
X bar (-/+) E
E = zc * ( s / sqrt (n))
= 1.96 * ( 74.5/sqrt(64))
= 18.2525
252.45 (-/+) 18.2525
234.20 and 270.70
The 95% confidence interval is 234.2 and 270.7
Part b)
The value of 215.60 does not fall in the above confidence interval. It appears that the population mean amount spent per day by families visiting Niagara Falls differs from the means reported by the American Automobiles.