A random sample of 42 business analysts at a management consulting firm revealed that they worked an...
A random sample of 42 business analysts at a management consulting firm revealed that they worked an average of 5.5 years on the job before being promoted to the associate level. The population standard deviation was 1.1 years. Using the 0.99 degree of confidence, what is the confidence interval for the population mean?
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A random sample of 42 business analysts at a
management consulting firm revealed that they worked an...
Next Page Previous Page Page 3 of 10 Question 3 (1 point) A random sample of...
Next Page Previous Page Page 3 of 10 Question 3 (1 point) A random sample of 85 group leaders and supervisors revealed that they worked an average of 6.5 years before being promoted. The population standard deviation was 1.7 years. What is the lower confidence limit for a 95% confidence interval? Express your solution rounded to two dedmal places (do not include symbols such as $ or %). Your Answer: Answer Page 3 of 10 Previous Page Next Page Submit...
A random sample of 85 group leaders, supervisors and similar personnel at General Motors revealed that,...
A random sample of 85 group leaders, supervisors and similar personnel at General Motors revealed that, on average, they spent 6.5 years in a particular job before being promoted. The standard deviation of the sample was 1.7 years. Construct a 95% confidence interval. Explain which words in the problem indicate a z-distribution or a t-distribution (choose one). Show how to calculate the confidence interval using Excel or typing out the equation (such as for proportions). The purpose is to identify...
A random sample of 13 customers in a Kentucky fast food restaurant revealed an average bill...
A random sample of 13 customers in a Kentucky fast food
restaurant revealed an average bill of OMR 18.75 per person. The
population standard deviation is OMR5.4. Estimate 98%
confidence
interval for the mean bill of all customers.
A random sample of ' n' customers in a Kentucky fast food restaurant revealed an average bill of OMR & per person. The population standard deviation is OMR 6. Estimate 98% confidence interval for the mean bill of all customers. n X...
A survey of a simple random sample of 140 dieters revealed that the numbers of times...
A survey of a simple random sample of 140 dieters revealed that the numbers of times they “cheated” on their diets had a mean of 7.0 times per week. Construct and interpret a 99% confidence interval for the mean number of times dieters “cheat” on their diets each week. Assume that the population standard deviation is 1.5 times per week.
Question 9 (1 point) For a random sample of 18 recent business school graduates beginning their first job, the mean...
Question 9 (1 point) For a random sample of 18 recent business school graduates beginning their first job, the mean starting salary was found to be $43,500, and the sample standard deviation was $5,500. Assuming the population is normally distributed, find the margin of error of a 95% confidence interval for the population mean.
5. A random sample of 625 customers of the restaurant xyz revealed the average monthly expenses...
5. A random sample of 625 customers of the restaurant xyz revealed the average monthly expenses in the amount of € 43.50 with a standard deviation of 19.20. a) Assuming that the dining expenses are normally distributed, construct a 95% confidence interval for the population mean expenses among xyz consumers( keep 2 decimals in your answer) b) How confident are you that the estimated 43.50 will be within 1.50 of the population mean? c) What would your margin of error...
A random sample of 2597 adult Americans in November 2009 revealed a sample mean of $...
A random sample of 2597 adult Americans in November 2009 revealed a sample mean of $ 343.31 that was spent on Christmas shopping over the Thanksgiving weekend a. Is $ 343.31 a parameter or a statistic? What symbol is used for this? The value $ 343.31 is a The symbol used is b. Determine the value of for a 95% confidence interval for the population mean amount spent on Christmas shopping over that weekend. c. What further information about the...
Management at a marketing consulting firm in Minnesota is concerned that many of their associates have...
Management at a marketing consulting firm in Minnesota is concerned that many of their associates have trouble getting to work during the winter season due to snow storms and other adverse weather conditions. An analysis by the HR department finds that most of the associates could actually perform their jobs without being in the office so long as they have a laptop and Internet access. Management would like to offer a work option to its associates to address this situation....
1. A random sample of 82 customers, who visited a department store, spent an average of...
1. A random sample of 82 customers, who visited a department store, spent an average of $71 at this store. Suppose the standard deviation of expenditures at this store is O = $19. What is the e 98% confidence interval for the population mean? 2. A sample of 25 elements produced a mean of 123.4 and a standard deviation of 18.32 Assuming that the population has a normal distribution, what is the 90% confidence interval for the population mean? 3....
For a random sample of 16 recent business school graduates beginning their first job, the mean...
For a random sample of 16 recent business school graduates beginning their first job, the mean starting salary was found to be $39,500, and the sample standard deviation was $8,500. Assuming the population is normally distributed, calculate the lower confidence limit (LCL) of the population mean with a 0.05.
Next Page Previous Page Page 3 of 10 Question 3 (1 point) A random sample of 85 group leaders and supervisors revealed that they worked an average of 6.5 years before being promoted. The population standard deviation was 1.7 years. What is the lower confidence limit for a 95% confidence interval? Express your solution rounded to two dedmal places (do not include symbols such as $ or %). Your Answer: Answer Page 3 of 10 Previous Page Next Page Submit...
A random sample of 13 customers in a Kentucky fast food
restaurant revealed an average bill of OMR 18.75 per person. The
population standard deviation is OMR5.4. Estimate 98%
confidence
interval for the mean bill of all customers.
A random sample of ' n' customers in a Kentucky fast food restaurant revealed an average bill of OMR & per person. The population standard deviation is OMR 6. Estimate 98% confidence interval for the mean bill of all customers. n X...
Question 9 (1 point) For a random sample of 18 recent business school graduates beginning their first job, the mean starting salary was found to be $43,500, and the sample standard deviation was $5,500. Assuming the population is normally distributed, find the margin of error of a 95% confidence interval for the population mean.
5. A random sample of 625 customers of the restaurant xyz revealed the average monthly expenses in the amount of € 43.50 with a standard deviation of 19.20. a) Assuming that the dining expenses are normally distributed, construct a 95% confidence interval for the population mean expenses among xyz consumers( keep 2 decimals in your answer) b) How confident are you that the estimated 43.50 will be within 1.50 of the population mean? c) What would your margin of error...
A random sample of 2597 adult Americans in November 2009 revealed a sample mean of $ 343.31 that was spent on Christmas shopping over the Thanksgiving weekend a. Is $ 343.31 a parameter or a statistic? What symbol is used for this? The value $ 343.31 is a The symbol used is b. Determine the value of for a 95% confidence interval for the population mean amount spent on Christmas shopping over that weekend. c. What further information about the...
Management at a marketing consulting firm in Minnesota is concerned that many of their associates have trouble getting to work during the winter season due to snow storms and other adverse weather conditions. An analysis by the HR department finds that most of the associates could actually perform their jobs without being in the office so long as they have a laptop and Internet access. Management would like to offer a work option to its associates to address this situation....
1. A random sample of 82 customers, who visited a department store, spent an average of $71 at this store. Suppose the standard deviation of expenditures at this store is O = $19. What is the e 98% confidence interval for the population mean? 2. A sample of 25 elements produced a mean of 123.4 and a standard deviation of 18.32 Assuming that the population has a normal distribution, what is the 90% confidence interval for the population mean? 3....
For a random sample of 16 recent business school graduates beginning their first job, the mean starting salary was found to be $39,500, and the sample standard deviation was $8,500. Assuming the population is normally distributed, calculate the lower confidence limit (LCL) of the population mean with a 0.05.
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