a. For 95% confidence interval, Z*-value = 1.96
Margin of safety = Z* x (standard deviation) / sqrt (sample) = 1.96 x (15 / sqrt (100)) = 2.94
b. Confidence interval = (Mean - Margin of error) to (mean + margin of error) = (50 - 2.94) to (50+2.94) = [$47.06, $52.94]
A sample survey of 54 discount brokers showed that the mean price charged for a trade...
A sample survey of 62 discount brokers showed that the mean price charged for a trade of 100 shares at $50 per share was $31.22. The survey is conducted annually. With the historical data available, assume a known population standard deviation of $19. (a) Using the sample data, what is the margin of error in dollars associated with a 95% confidence interval? (Round your answer to the nearest cant) $ (b) Develop a 95% confidence interval for the mean price...
A sample survey of 59 discount brokers showed that the mean price charged for a trade of 100 shares at $50 per share was $33.71. The survey is conducted annually. With the historical data available, assume a known population standard deviation of $13. a. Using the sample data, what is the margin of error associated with a 90% confidence interval (to 2 decimals)? b. Develop a confidence interval for the mean price charged by discount brokers for a trade of 100...
1. A sample survey of 54 discount brokers showed that the mean price charged for a trade of 100 shares at $50 per share was S33.77 and a standard deviation of S15. Develop a 95% confidence interval for the mean price charged by discount brokers for a trade of 100 shares at $50 per share. a) 6 marks b) Explain what the interval you found tells you 2 marks e) What sample size would be necessary to achieve a margin...
A survey of 54 randomly selected "iPhone" owners showed that the purchase price has a mean of $400 with a sample standard deviation of $170. a. Compute the standard error of the sample mean. (Round the final answer to the nearest whole number.) Standard error b. Compute the 80% confidence interval for the mean. (Round the final answers to 2 decimal places.) The confidence interval is between $ and $ . c. How large a sample is needed to estimate...
A survey of 41 randomly selected "iPhone" owners showed that the purchase price has a mean of $414 with a sample standard deviation of $180. a. Compute the standard error of the sample mean. (Round the final answer to the nearest whole number.) Standard error b. Compute the 80% confidence interval for the mean. (Round the final answers to 2 decimal places.) The confidence interval is between $ and $ . c. How large a sample is needed to estimate...
A sample of 100 units showed a sample mean 12 oz with a standard deviation 3 oz. Obtain the 95% confidence interval for the population mean. Question options: [11.2 , 12.8] [11.4 , 12.6] [11.1 , 12.9] [11.6 , 12.6]
A survey conducted by the American Automobile Association showed that a family of four spends an average of $215.60 per day while on vacation Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per day and a sample standard deviation of $79.50 Devo a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls (to 2 decimals). . Based on the...
Supplementary Exercises 385 A survey conducted by the American Automobile Association (AAA) showed that a fam- ily of four spends an average of $215.60 per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per 45. day and a sample standard deviation of $74.50 Develop a 95% confidence interval estimate of the mean amount spent per day by a a. family of four visiting Niagara Falls....
An article reported that for a sample of 54 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO2 level (ppm) was 654.16, and the sample standard deviation was 162. a) Calculate and interpret a 95% two sided answers to two decimal places.) confidence interval for true average CO2 level n the population of a homes from hich the sample as selected Round your ppm Interpret the resulting interval o we are 95% confident that the...
Question 4 1 pts A survey of 400 students provides a sample mean of 7.10 hours worked per week. From previous studies, the researcher knows that the standard deviation of this variable is 5 hours. What is a 95% confidence interval based on this sample? (6.10, 8.10) (6.45, 7.75) (6.61, 7.59) (7.10, 8.48)