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a growing seasons for a random sample of 31 population standard deviation is 31.8 days, 05,...
Suppose a random sample of 36 is selected from a population with a standard deviation of...
Suppose a random sample of 36 is selected from a population with a standard deviation of 12. If the sample mean is 98, the 99% confidence Interval to estimate the population mean is what.
A statistics practitioner took a random sample of 54 observations from a population whose standard deviation...
A statistics practitioner took a random sample of 54 observations from a population whose standard deviation is 33 and computed the sample mean to be 105. Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits. A. Estimate the population mean with 95% confidence. Confidence Interval = B. Estimate the population mean with 90% confidence. Confidence interval = C. Estimate the population mean with 99%...
A random sample of 28 items is drawn from a population whose standard deviation is unknown....
A random sample of 28 items is drawn from a population whose standard deviation is unknown. The sample mean is x = 790 and the sample standard deviation is s=15. Use Appendix D to find the values of Student'st (a) Construct an interval estimate of u with 99% confidence. (Round your answers to 3 decimal places.) The 99% confidence interval is from 780:36 to 737854 (b) Construct an interval estimate of u with 99% confidence, assuming that s- 30. (Round...
The days of training a new employee needs are normally distributed with a population standard deviation...
The days of training a new employee needs are normally distributed with a population standard deviation of 3 days and an unknown population mean. If a random sample of 23 new employees is taken and results in a sample mean of 18 days, use Excel to find a 90% confidence interval for the population mean. Round the final answer to two decimal places.
You are given the sample mean and the population standard deviation. Use this information to construct...
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. From a random sample of 49 business days, the mean closing price of a certain stock was $113.41. Assume the population standard deviation is $11.13. a) The 90% confidence interval is (?) - (?)
You are given the sample mean and the population standard deviation. Use this information to construct...
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. From a random sample of 39 business days, the mean closing price of a certain stock was $113.67. Assume the population standard deviation is $10.91. The 90% confidence interval is (1 , b). (Round to two decimal places as needed.)
13. A random sample is drawn from a population of known standard deviation 11.3. Construct a...
13. A random sample is drawn from a population of known standard deviation 11.3. Construct a 90% confidence interval for the population mean based on the information given (not all of the information given need be used). a. n-36, x =105.2, s=11.2 b. n-100, x =105.2, s - 11.2 14. A random sample is drawn from a population of unknown standard deviation. Construct a 99%99% confidence interval for the population mean based on the information given. a. n=49, x 17.1,...
2. A population is known to have a standard deviation of 26.1. A sample space of...
2. A population is known to have a standard deviation of 26.1. A sample space of 35 items has a mean of (1 point) 562. Construct a 90% confidence interval estimate of the mean of the population. 0566<p<558 0555<pバ569 O 551<H573 0561<p<563 3. While researching the cost of school lunches per week across the state, you use a sample size of 45(point) weekly lunch prices. The standard deviation is known to be 68 cents. In order to be 90% confident,...
A simple random sample of size 14 has mean 3.67 and standard deviation 1.75. The population...
A simple random sample of size 14 has mean 3.67 and standard deviation 1.75. The population is approximately normally distributed. Construct a 99% confidence interval for the population mean. 1. The parapmeter is the population (choose one) mean, standard deviation, variable, proportion 2. The correct method to find the confidence interval is the (choose one) z, t, chi square method
2. A random sample of 25 watermelons from New Seasons were weighed, generating a sample mean...
2. A random sample of 25 watermelons from New Seasons were weighed, generating a sample mean of xã = 21 pounds with a standard deviation of s = 3 pounds. a) Construct a 98% confidence interval for the population mean weight of watermelons at New Seasons, b) Interpret: c) New Seasons claims that their watermelons have an average weight of 22 pounds. Does your sample from above provide sufficient evidence that the population mean of watermelons is actually less than...
A statistics practitioner took a random sample of 54 observations from a population whose standard deviation is 33 and computed the sample mean to be 105. Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits. A. Estimate the population mean with 95% confidence. Confidence Interval = B. Estimate the population mean with 90% confidence. Confidence interval = C. Estimate the population mean with 99%...
A random sample of 28 items is drawn from a population whose standard deviation is unknown. The sample mean is x = 790 and the sample standard deviation is s=15. Use Appendix D to find the values of Student'st (a) Construct an interval estimate of u with 99% confidence. (Round your answers to 3 decimal places.) The 99% confidence interval is from 780:36 to 737854 (b) Construct an interval estimate of u with 99% confidence, assuming that s- 30. (Round...
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. From a random sample of 39 business days, the mean closing price of a certain stock was $113.67. Assume the population standard deviation is $10.91. The 90% confidence interval is (1 , b). (Round to two decimal places as needed.)
13. A random sample is drawn from a population of known standard deviation 11.3. Construct a 90% confidence interval for the population mean based on the information given (not all of the information given need be used). a. n-36, x =105.2, s=11.2 b. n-100, x =105.2, s - 11.2 14. A random sample is drawn from a population of unknown standard deviation. Construct a 99%99% confidence interval for the population mean based on the information given. a. n=49, x 17.1,...
2. A population is known to have a standard deviation of 26.1. A sample space of 35 items has a mean of (1 point) 562. Construct a 90% confidence interval estimate of the mean of the population. 0566<p<558 0555<pバ569 O 551<H573 0561<p<563 3. While researching the cost of school lunches per week across the state, you use a sample size of 45(point) weekly lunch prices. The standard deviation is known to be 68 cents. In order to be 90% confident,...
2. A random sample of 25 watermelons from New Seasons were weighed, generating a sample mean of xã = 21 pounds with a standard deviation of s = 3 pounds. a) Construct a 98% confidence interval for the population mean weight of watermelons at New Seasons, b) Interpret: c) New Seasons claims that their watermelons have an average weight of 22 pounds. Does your sample from above provide sufficient evidence that the population mean of watermelons is actually less than...
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