Exercise 9-61 (LO9-2, LO9-4) A survey of 48 randomly selected iPhone owners showed that the purchase...
Exercise 9-61 (LO9-2, LO9-4) A survey of 48 randomly selected iPhone owners showed that the purchase price has a mean of $670 with a sample standard deviation of $23. (Use z Distribution Table.)
Compute the standard error of the sample mean. (Round your answer to the nearest whole number.)
Standard error:
Compute the 98% confidence interval for the mean. (Use t Distribution Table.) (Round your answers to 3 decimal places.)
The confidence interval is between: ________ and ________
To be 98% confident, how large a sample is needed to estimate the population mean within $8? (Round up your answer to the next whole number.)
Sample size:
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Exercise 9-61 (LO9-2, LO9-4) A survey of 48 randomly selected
iPhone owners showed that the purchase...
A survey of 41 randomly selected "iPhone" owners showed that the purchase price has a mean...
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 survey of 54 randomly selected "iPhone" owners showed that the purchase price has a mean...
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...
er 710 9 You skipped this question in the previous attempt Exercise 9-44 (LO9-1, LO9-2) PrintTech,...
er 710 9 You skipped this question in the previous attempt Exercise 9-44 (LO9-1, LO9-2) PrintTech, Inc is introducing a new line of ink-jet printers and would like to promote the number of pages a user can expect from a print cartridge. A sample of 10 cartridges revealed the following number of pages printed (Use Distribution Table) 2,983 3,96 2,518 2,559 2,559 2,622 2,537 2,485 3,236 2,758 Click here for the Excel Deta File a. What is the point estimate...
A survey of 26 randomly sampled judges employed by the state of Florida found that they...
A survey of 26 randomly sampled judges employed by the state of Florida found that they earned an average wage (including benefits) of $61.00 per hour. The sample standard deviation was $6.08 per hour. (Use t Distribution Table.) a.What is the best estimate of the population mean?________. b.Develop a 90% confidence interval for the population mean wage (including benefits) for these employees. (Round your answers to 2 decimal places.) _______ and_________ c.How large a sample is needed to assess the...
33 46 48 21 30 32 10 31 20 29 A survey of 25 randomly selected...
33 46 48 21 30 32 10 31 20 29 A survey of 25 randomly selected customers found the ages shown (in years). The mean is 31.76 years and the standard deviation is 10.37 years. a) Construct a 98 % confidence interval for the mean age of all customers, assuming that the assumptions and conditions for the confidence 20 38 40 35 40 interval have been met. 22 37 10 41 26 b) How large is the margin of error?...
A sample of 14 small bags of the same brand of candies was selected. Assume that...
A sample of 14 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is normal. The weight of each bag was then recorded. The mean weight was 3 ounces with a standard deviation of 0.15 ounces. The population standard deviation is known to be 0.1 ounce. NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that...
A sample survey of 62 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...
Last year, 48% of business owners gave a holiday gift to their employees. A survey of...
Last year, 48% of business owners gave a holiday gift to their employees. A survey of business owners conducted this year indicates that 37% planned to provide a holiday gift to their employees. Suppose the survey results are based on a sample of 60 business owners. a. How many business owners in the survey planned to provide a holiday gift to their employees this year? Round your answer to the nearest whole number. b. Suppose the business owners in the...
1,600 college freshmen were randomly selected for a national survey. Among survey participants, the mean grade-point...
1,600 college freshmen were randomly selected for a national survey. Among survey participants, the mean grade-point average (GPA) was 3 , and the margin of error was 0.0343% for the 95% confidence interval. If we want to get a 85% confidence interval with a small margin of error 15%. How many students do you need to select? Remember that our answer is about actual people -- to answer this question you must round your number up to the nearest whole...
A random sample of 88 women showed that the mean number of children reported was 1.58...
A random sample of 88 women showed that the mean number of children reported was 1.58 with a standard deviation of 1.8. (interestingly, a sample of 88 men showed a mean of 1.3 children.) Complete parts (a) through (c) below EEB Click the lcon to view the ttable. Find a 95% confidence interval for the population mean number of children for women. Because the sample size is so large, you can use 1.96 for the critical value of t (which...
er 710 9 You skipped this question in the previous attempt Exercise 9-44 (LO9-1, LO9-2) PrintTech, Inc is introducing a new line of ink-jet printers and would like to promote the number of pages a user can expect from a print cartridge. A sample of 10 cartridges revealed the following number of pages printed (Use Distribution Table) 2,983 3,96 2,518 2,559 2,559 2,622 2,537 2,485 3,236 2,758 Click here for the Excel Deta File a. What is the point estimate...
33 46 48 21 30 32 10 31 20 29 A survey of 25 randomly selected customers found the ages shown (in years). The mean is 31.76 years and the standard deviation is 10.37 years. a) Construct a 98 % confidence interval for the mean age of all customers, assuming that the assumptions and conditions for the confidence 20 38 40 35 40 interval have been met. 22 37 10 41 26 b) How large is the margin of error?...
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...
Last year, 48% of business owners gave a holiday gift to their employees. A survey of business owners conducted this year indicates that 37% planned to provide a holiday gift to their employees. Suppose the survey results are based on a sample of 60 business owners. a. How many business owners in the survey planned to provide a holiday gift to their employees this year? Round your answer to the nearest whole number. b. Suppose the business owners in the...
1,600 college freshmen were randomly selected for a national survey. Among survey participants, the mean grade-point average (GPA) was 3 , and the margin of error was 0.0343% for the 95% confidence interval. If we want to get a 85% confidence interval with a small margin of error 15%. How many students do you need to select? Remember that our answer is about actual people -- to answer this question you must round your number up to the nearest whole...
A random sample of 88 women showed that the mean number of children reported was 1.58 with a standard deviation of 1.8. (interestingly, a sample of 88 men showed a mean of 1.3 children.) Complete parts (a) through (c) below EEB Click the lcon to view the ttable. Find a 95% confidence interval for the population mean number of children for women. Because the sample size is so large, you can use 1.96 for the critical value of t (which...
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