In a random sample of 8 people, the mean commute time to work was 35.5 minute and the standard deviation was 7.3 minutes. A 95% confidence interval using t-distribution was calculated to be (29.4, 41.6). 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 95% confidence interval using the standard normal distribution with the appropriate calculations for standard
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In a random sample of 8 people, the
A 95% confidence interval using t-distribution was calculated to be (29.4, 41.6). 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 95% confidence interval using the standard
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- 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 98% confidence interval using the t-distribution was calculated to be (28.7,44.3). After researching commute times to work, it was found that the population standard deviation is 8.7 minutes. Find the margin of error and construct a 98% confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare the results. The margin of error of μ is 7.27.2. A 98% confidence interval using the standard normal distribution is ,______.In a random sample of 8 people, the mean commute time to work was 35.5 minutes and the standard deviation was 7.4 minutes. A 98% confidence interval using the t-distribution was calculated to be (27.7,43.3). After researching commute times to work, it was found that the population standard deviation is 8.8 minutes. Find the margin of error and construct a 98% confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare the results. The margin of error of u is (Round to two decimal places as needed.) ...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 90% confidence interval using the t-distribution was calculated to be (29.6,39.4). After researching commute times to work, it was found that the population standard deviation is 9.7 minutes. Find the margin of error and construct a 90% confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare the results. The margin of error of μ is ___ A 90% confidence interval using the standard normal distribution is ( __ , __ )
- In a random sample of 8 people, the mean commute time to work was 36.5 minutes and the standard deviation was 7.2 minutes. A 90% confidence interval using the t-distribution was calculated to be (31.7,41.3). 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 the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare the results. The margin of error of u is (Round to two decimal places as needed.)In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 5.5 and a standard deviation of 19.8. Construct a 99% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol? Click here to view at distribution table. Click here to view page 1 of the standard normal distribution table, Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean u? O mg/dL < µIn a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 3.9 and a standard deviation of 15.3. Construct a 99% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol? Click here to view at distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean u? | mg/dL < µA clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 20 subjects had a mean wake time of 105.0 min. After treatment, the 20 subjects had a mean wake time of 82.5 min and a standard deviation of 22.5 min. Assume that the 20 sample values appear to be from a normally distributed population and construct a 90% confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 105.0 min before the treatment? Does the drug appear to be effective? Construct the 90% confidence interval estimate of the mean wake time for a population with the treatment. ???? min<μ<???? minIn 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 95% confidence interval using the t-distribution was calculated to be (30.3,42.7). 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 the results. The margin of error of u is (Round to two decimal places as needed.)In a sample of 10 randomly selected black bears from northeast Pennsylvania, the sample mean was 214.1 pounds and the sample standard deviation was 51.1 pounds. Construct a 99% confidence interval estimate for the population standard deviation of all black bears in this region.In a random sample of 8 people, the mean commute time to work was 34.5 minutes and the standard deviation was 7.2 minutes. A 90% confidence interval using the t-distribution was calculated to be (29.7,39.3). After researching commute times to work, it was found that the population standard deviation is 8.8 minutes. Find the margin of error and construct a 90% confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare the results. C The margin of error of μ is. (Round to two decimal places as needed.) A 90% confidence interval using the standard normal distribution is (0. (Round to one decimal place as needed.) Compare the results. Choose the correct answer below. O A. The confidence interval found using the standard normal distribution has smaller lower and upper confidence interval limits. OB. The confidence interval found using the standard normal distribution is the same as the confidence interval…A small university is trying to monitor its electricity usage. For a random sample of 30 weekend days (Saturdays and Sundays), the student center used an average of 94.26 kilowatt hours (kWh) with standard deviation 43.29. For a random sample of 60 weekdays, (Monday - Friday), the student center used an average of 112.63 kWh with standard deviation 32.07. Construct a 95% confidence interval for the difference in mean electricity use at the student center between weekdays and weekend days. Use two decimal places in your margin of error. -6.80 to 43.54 kWh -4.61 to 41.35 kWh -3.38 to 40.51 kWh 0.12 to 36.62 kWhA sample of 120 students yielded a sample mean GPA of 3.35, end a sample standard deviation equal to 0.28. 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