In a survey relating to the mean time for a drug to take effect, a sample...
In a survey relating to the mean time for a drug to take effect, a sample of 400 trials showed the mean time of 26 minutes with standard deviation of 4 minutes. Find the minimum and maximum “usual” times. Is 10 minutes Usual?
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In a survey relating to the mean time for a drug to take effect,
a sample...
A survey of a random sample of 1000 smartphone owners found that the mean daily time...
A survey of a random sample of 1000 smartphone owners found that the mean daily time spent communicating on a smart phone was 131.4 minutes if we assume the population standard deviation is 21.2 minutes the 95% confidence interval for the population mean will be 130.1<u<132. 7minutes. what is the meaning of these results
A survey is taken on average commute time to work. The mean and standard deviation of...
A survey is taken on average commute time to work. The mean and standard deviation of the survey are as follows: Mean commute time: 30 minutes Standard deviation: 10 minutes 14. Use the mean and standard deviation above to calculate the z-score of a 45 minute commute to work. Explain what this z-score tells us.
A survey across banks in Delhi showed the mean time taken to serve customers was 6.4...
A survey across banks in Delhi showed the mean time taken to serve customers was 6.4 minutes while the standard deviation was 1.5 minutes. It is not known whether the data distribution is symmetrical. a) What is the probability of a customer being served between 2.8 and 10 minutes? b) What is the probability of a customer being served at the expected value of the distribution? Show the sketches for your answers.
A new drug is introduced that is supposed to reduce fevers. Tests are done with the...
A new drug is introduced that is supposed to reduce fevers. Tests are done with the drug. The drug is given to 70 people who have fevers. It is found that the mean time that it takes for the fever to get back to normal for this test group is 400 minutes with a standard deviation of 85 minutes. Find the 95% confidence interval for the mean time that the drug will take to reduce all fevers for all people.
The director of research and development is testing a new drug. She wants to know if there is evidence at the 0.05 level that the drug stays in the system for more than 366 minutes. For a sample of 47patients, the mean time the drug stayed in the system w
The director of research and development is testing a new drug. She wants to know if there is evidence at the 0.05 level that the drug stays in the system for more than 366 minutes. For a sample of 47patients, the mean time the drug stayed in the system was 371 minutes. Assume the population standard deviation is 25. Make the decision to reject or fail to reject the null hypothesis.
Suppose a survey of 36 people determined that the sample mean time spent on the internet...
Suppose a survey of 36 people determined that the sample mean time spent on the internet was 8.2 hours with a sample standard deviation of 9.8 hours. What is the critical value for a 95% confidence interval? Note: Enter X.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 7 is entered as 7.00, 3.5 is entered as 3.50, 0.3750 is entered as 0.38 | | Suppose a survey of 36 people determined that the sample...
in a random sample of 8 people, the mean commute time to work was 36.5 minutes...
in a random sample of 8 people, the mean commute time to work was 36.5 minutes and the standard deviation was 7.4 minutes. A 90% confidence interval using the t-distribution was calculated to be (31.5,41.5). After researching commute times to work, it was found that the population standard deviation is 9.3 minutes. find the margin of error and construct a 90% confidence interval using normal distribution with the appropriate calculations for a standard deviation that is known. compare the results....
In a random sample of 8 people, the mean commute time to work was 34.5 minutes...
In a random sample of 8 people, the mean commute time to work was 34.5 minutes and the standard deviation was 7.3 minutes. A 95% confidence interval using the t-distribution was calculated to be (28.4.40.6). After researching commute times to work, it was found that the population standard deviation is 9.4 minutes. Find the margin of error and construct a 95% confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare...
In a random sample of 88 people, the mean commute time to work was 36.536.5 minutes...
In a random sample of 88 people, the mean commute time to work was 36.536.5 minutes and the standard deviation was 7.37.3 minutes. A 9090% confidence interval using the t-distribution was calculated to be left parenthesis 31.6 comma 41.4 right parenthesis(31.6,41.4). After researching commute times to work, it was found that the population standard deviation is 9.39.3 minutes. Find the margin of error and construct a 9090% confidence interval using the standard normal distribution with the appropriate calculations for a...
In a random sample of 88 people, the mean commute time to work was 35.5 minutes...
In a random sample of 88 people, the mean commute time to work was 35.5 minutes and the standard deviation was 7.2 minutes. A 90% confidence interval using the t-distribution was calculated to be left parenthesis 30.7 comma 40.3 right parenthesis(30.7,40.3). After researching commute times to work, it was found that the population standard deviation is 8.9 minutes. Find the margin of error and construct a 90% confidence interval using the standard normal distribution with the appropriate calculations for a...
In a random sample of 8 people, the mean commute time to work was 34.5 minutes and the standard deviation was 7.3 minutes. A 95% confidence interval using the t-distribution was calculated to be (28.4.40.6). After researching commute times to work, it was found that the population standard deviation is 9.4 minutes. Find the margin of error and construct a 95% confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare...
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