In a study of factors affecting hypnotism, visual analogue scale (VAS) sensory ratings were 16 subjects. For these sample ratings, the mean is 8.33 with a standard deviation of 1.96. Co 99% confidence interval for the true mean sensory ratings
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- On average, adults spend approximately 5 hours per night in NREM (non-rapid eye movement) sleep. A sleep study selects a random sample of adults and gives them each a weighted blanket. They would like to know if using the weighted blanket changes the mean number of hours per night of NREM sleep. We want to test = 5 versus ≠ 5 where = the true mean hours of NREM sleep per night for adults who use a weighted blanket. A 99% confidence interval for the true mean hours of NREM sleep per night for adults who use a weighted blanket is (5.1, 5.8) hours. Based on the confidence interval, what conclusion would you make for a test of these hypotheses? The 99% confidence interval include as a plausible value, so we would H0. We convincing evidence that the true mean hours of NREM sleep per night for adults who use a weighted blanket 5 hours.Caffeine is considered the most commonly used psychoactive drug in the world. It is estimated that the average daily caffeine consumption among 120 surveyed randomly selected Filipinos is about 280 mg per day with a standard deviation of 10 mg. Construct a 90% confidence interval for the true mean daily caffeine consumption of Filipinos. Interpret the results.Sales personnel for City Distributors Inc. submit weekly reports listing the customer complaints made during the week. A simple random sample of 8 weekly reports showed a sample mean of 8 complaints per week with a sample standard deviation of 3.464. What is the 95% confidence interval for the population mean?
- Intellectual development (Perry) scores were determined for 21 students in a first-year, project-based design course. (Recall that a Perry score of 1 indicates the lowest level of intellectual development, and a Perry score of 5 indicates the highest level.) The average Perry score for the 21 students was 3.27 and the standard deviation was .40. Apply the confidence interval method to estimate the true mean Perry score of all undergraduate engineering students with 90% confidence. Interpret the results.Please share the solution for the last 2 parts (c & d) in bold. The first three parts answered by your team. In a study of the Gouldian finch, Griffith et al. (2011) looked at stress caused by having an incompatible mate.There are two genetically distinct types of Gouldian finches, one having a red face and the other having a black face. Previous experiments have shown that female finches have a strong preference for mating with males with the same face color as themselves , and that when different face types of finch mate with one another, their offspring are less likely to survive than when both parents are the same type. Researchers paired females sequentially with males of both types in random order,In other words, each female bred twice once with a compatible male and once with an incompatible male.Each time, females produced a brood of young with the assigned male.For each pairing, the researchers measured the blood corticosterone concentration (in units of ng/ml) as an index…HOLY BIBLE College tuition: A simple random sample of 35 Colleges and universities in the united states had a mean tuition of $18,702 with a standard deviation of $10,653. Construct a 95% Confidence interval for the mean tuition for all Colleges and Universities in the united states.
- In a test of the effectiveness of garlic for lowering cholesterol, 64 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.4 and a standard deviation of 23.9. Use a 0.05 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. What are the null and alternative hypotheses? Ο Α. Ho: μ=0 mg/ldL H1: µ > 0 mg/dL B. Ho : μ=0 mg/dL H1: µ0 mg/dL H1: µ<0 mg/dL D. Ho: µ=0 mg/dL H1:µ#0 mg/dLIn a test of the effectiveness of garlic for lowering cholesterol, 36 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.7 and a standard deviation of 2.33. Use a 0.01 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.Fuel efficiency of Prius: Fueleconomy.gov, the official US government source for fuel economy information, allows users to share gas mileage information on their vehicles. The histogram below shows the distribution of gas mileage in miles per gallon (MPG) from 14 users who drive a 2012 Toyota Prius. The sample mean is 53.3 MPG and the standard deviation is 5.2 MPG. Note that these data are user estimates and since the source data cannot be verified, the accuracy of these estimates are not guaranteed. (a) The EPA claims that a 2012 Prius gets 50 MPG (city and highway mileage combined). Do these data provide strong evidence against this estimate for drivers who participate on fueleconomy.gov? The test statistic is: __________________(please round to two decimal places)The p-value:___________________________ for this hypothesis test is: (please round to four decimal places)(c) Please comment on the primary thing that is wrong about the following statement: There is sufficient evidence…
- 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?In a test of the effectiveness of garlic for lowering cholesterol, 64 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.6 and a standard deviation of 1.92. Use a 0.05 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. What are the null and alternative hypotheses?In a test of the effectiveness of garlic for lowering cholesterol, 64 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.9 and a standard deviation of 16.9 . Use a 0.01 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0 . What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. What are the null and alternative hypotheses? A. Upper H 0 : mu equals0 mg/dL Upper H 1 : mu greater than0 mg/dL B. Upper H 0 : mu greater than0 mg/dL Upper H 1 : mu less than0 mg/dL C. Upper H 0 : mu…