A clinical trial was conducted to test the effectiveness of a drug for treating insomniastandard deviation of 20.4 min. Assume that the 20 sample values appear to be from a normally distributed population and construct a 95% confidence interval estimate of the mean wake time for a population with drug treatments. What does theresult suggest about the mean wake time of 104.0 min before the treatment? Does the drug appear to be effective?older subjects. Before treatment, 20 subjects had a mean wake time of 104.0 min. After treatment, the 20 subjects had a mean wake tíme of 98.4 min and aConstruct the 95% confidence interval estimate of the meah wake time for a population with the treatment.min

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Answered: A clinical trial was conducted to test…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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In a randoms sample of six people, the mean driving distance to work was 19.7 miles and the standard deviation was 6.1 miles. Assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 90% confidence interval for the population mean u. Interpret the results.
The principal at a large high school claims that students spend at least 10 hours per week doing homework, on average. To investigate this claim, an AP Statistics class selected a random sample of 250 students from their school and asked them how long they spent doing homework during the last week. The sample mean was 10.2 hours and the sample standard deviation was 4.2 hours. Construct and interpret a 95% confidence interval for the mean time that students at this school spent doing homework in the last week. Based on your interval, what can you conclude about the principal's claim? What is your t* value? Remember to use the conservative DF value.
You were told that the mean daily cost of a hotel room in downtown St. Louis is $120. You will be needing to rent a hotel room there in the near future and you want to find the best rate. You randomly select a sample of 18 hotel rooms and find that the average cost is $121.33 with a standard deviation of $18.046. At the α = 0.05 significance level, determine if there is enough evidence to reject the claim. What is the decision for this hypothesis test? You were told that the mean daily cost of a hotel room in downtown St. Louis is $120. You will be needing to rent a hotel room there in the near future and you want to find the best rate. You randomly select a sample of 18 hotel rooms and find that the average cost is $121.33 with a standard deviation of $18.046. At the α = 0.05 significance level, determine if there is enough evidence to reject the claim. What is the decision for this hypothesis test? Reject the null because the test statistic is in the critical region and the…
A study is conducted to find the mean life of a certain manufacturer's car batteries. A random sample of 34 batteries is selected and a mean of 47.5 months with a standard deviation of 7.3 months is found. Find a 95% confidence interval for the mean battery life.
In a random sample of 5 mobile devices, the mean repair cost was $60.00 and the standard deviation was $12.00. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a 95% confidence interval for the population mean. Interpret the results.
The average gold content from an ore sample in 36 different locations is found to be 2.6 grams per ton. a. Find the 95% and 99% confidence intervals for the mean gold content in that area. Assume that the population variance is 0.09.
A wildlife study is designed to find the mean weight of salmon caught by an Alaskan fishing company. Find the minimum sample size needed in order to construct a 95% confidence interval for the mean weight of such salmon, to within 0.30 pounds. Assume that the standard deviation of 3.10.
A study was conducted to estimate hospital costs for car accident victims who woreseat belts. Suppose that 46 randomly selected cases had an average of $15150and a standard deviation of $8500. If the average hospital costs for those that didnot wear seatbelts was $19000, construct a 90% confidence interval to see if theaverage costs are significantly different.
A sample of 12 suburban hotels showed that the mean revenue is $1117. It is known that the standard deviation is $121.8. Construct the 90% confidence interval for the true average revenue of all suburban hotels.
in a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes in their levels of LDL cholesterol have a mean of 5.7 and a standard deviation of 17.7. 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 and reducing LDL cholesterol?
The price of a share of stock divided by the company's estimated future earnings per share is called the P/E ratio. High P/E ratios usually indicate "growth" stocks, or maybe stocks that are simply overpriced. Low P/E ratios indicate "value" stocks or bargain stocks. A random sample of 51 of the largest companies in the United States gave the following P/E ratios†. 11 35 19 13 15 21 40 18 60 72 9 20 29 53 16 26 21 14 21 27 10 12 47 14 33 14 18 17 20 19 13 25 23 27 5 16 8 49 44 20 27 8 19 12 31 67 51 26 19 18 32 (a) Use a calculator with mean and sample standard deviation keys to find the sample mean x and sample standard deviation s. (Round your answers to four decimal places.) x = s = (b) Find a 90% confidence interval for the P/E population mean μ of all large U.S. companies. (Round your answers to one decimal place.) lower limit upper limit
After a sewer spill into a lake, an environmentalist wanted to test to see if the water was safe (mean bacteria count per unit volume of water is less than 200). Ten samples along different locations of the lake resulted in a sample mean of 192.7 and a sample standard deviation of 10.81. 2. a. Construct a 95% confidence interval for the mean.

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