Question

In: Statistics and Probability

A publisher wants to estimate the mean length of time​ (in minutes) all adults spend reading...

A publisher wants to estimate the mean length of time​ (in minutes) all adults spend reading newspapers. To determine this​ estimate, the publisher takes a random sample of 15 people and obtains the results below. From past​ studies, the publisher assumes sigma is 1.5 minutes and that the population of times is normally distributed. 10 9 10 6 10 8 9 8 12 11 11 10 12 9 12 Construct the​ 90% and​ 99% confidence intervals for the population mean. Which interval is​ wider? If​ convenient, use technology to construct the confidence intervals. The​ 90% confidence interval is

Solutions

Expert Solution

orchestra answered 2 years ago
Next > < Previous

Related Solutions

A publisher wants to estimate the mean length of time​ (in minutes) all adults spend reading...
A publisher wants to estimate the mean length of time​ (in minutes) all adults spend reading newspapers. To determine this​ estimate, the publisher takes a random sample of 15 people and obtains the results below. From past​ studies, the publisher assumes σ is 2.1 minutes and that the population of times is normally distributed. 10 12 7 10 11 8 12 8 11 7 6 11 8 8 12 a) The​ 90% confidence interval is ​____ b) The​ 99% confidence...
A research council wants to estimate the mean length of time (in minutes) that the average...
A research council wants to estimate the mean length of time (in minutes) that the average U.S. adult spends watching television using digital video recorders (DVR’s) each day. To determine the estimate, the research council takes random samples of 35 U.S. adults and obtains the following times in minutes. 24 27 26 29 33 21 18 24 23 34 17 15 19 23 25 29 36 19 18 22 16 45 32 12 24 35 14 40 30 19 14...
A research council wants to estimate the mean length of time (in minutes) that the average...
A research council wants to estimate the mean length of time (in minutes) that the average U.S. adult spends watching television using digital video recorders (DVR’s) each day. To determine the estimate, the research council takes random samples of 35 U.S. adults and obtains the following times in minutes. 24 27 26 29 33 21 18 24 23 34 17 15 19 23 25 29 36 19 18 22 16 45 32 12 24 35 14 40 30 19 14...
An advertising executive wants to estimate the mean amount of time that consumers spend with digital...
An advertising executive wants to estimate the mean amount of time that consumers spend with digital media daily. From past​ studies, the standard deviation is estimated as 42 minutes. A. What sample size is needed if the executive wants to be 90% confident of being correct to within plus or minus 4 minutes? B. If 99% confidence is desired, how many consumers need to be selected?
a hospital administrator wants to estimate the mean length of stay for all inpatients in the...
a hospital administrator wants to estimate the mean length of stay for all inpatients in the hospital. Based on a random sample of 676 patients from the previous year, she finds that the sample mean is 5.3 days with a standard deviation of 1.2 days. Construct and interpret a 95% and a 99% confidence interval for the mean.
An advertising executive wants to estimate the mean weekly amount of time consumers spend watching traditional...
An advertising executive wants to estimate the mean weekly amount of time consumers spend watching traditional television daily. Based on previous​ studies, the standard deviation is assumed to be 20 minutes. The executive wants to​ estimate, with ​99% confidence, the mean weekly amount of time to within ±5 minutes. a. What sample size is​ needed? b. If​ 95% confidence is​ desired, how many consumers need to be​ selected?
An advertising executive wants to estimate the mean weekly amount of time consumers spend watching traditional...
An advertising executive wants to estimate the mean weekly amount of time consumers spend watching traditional television daily. Based on previous​ studies, the standard deviation is assumed to be 21 minutes. The executive wants to​ estimate, with​ 99% confidence, the mean weekly amount of time to within ± 6 minutes. a. What sample size is​ needed? b. If​ 95% confidence is​ desired, how many consumers need to be​ selected?
An advertising executive wants to estimate the mean weekly amount of time 18- to 24-year-olds spend...
An advertising executive wants to estimate the mean weekly amount of time 18- to 24-year-olds spend watching traditional television in a large city. Based on studies in other cities, the standard deviation is assumed to be 10 minutes. The executive wants to estimate, within 99% confidence, the mean weekly amount of time within ±3 minutes. a. What sample size is needed? b. If 95% confidence is desired, how many 18- to 24-year-olds need to be selected? c. If she wants...
A group of researchers estimates the mean length of time (in minutes) the average U.S. adult...
A group of researchers estimates the mean length of time (in minutes) the average U.S. adult spends watching television using digital recorders and other forms of time-shifted television each day. To do so, the group takes a random sample of 40 U.S. adults and obtains the times (in minutes) listed below. Assume the population standard deviation to be 6.5 minutes. construct a 90% and 99% confidence interval for the population mean. 29, 27, 28, 26, 14,28,32, 12, 16, 21, 17,...
Suppose that Apple Inc. wished to estimate the mean time people spend on their phones every...
Suppose that Apple Inc. wished to estimate the mean time people spend on their phones every day. In a random sample of 30 phones users the mean time was found to be 150 minutes with a standard deviation of 70 minutes. * Construct a 90% confidence interval for the mean time people spend on their phone every day. * Construct a 95% confidence interval for the mean time people spend on their phone every day. * Construct a 99% confidence...
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT