A common claim is that garlic lowers cholesterol levels. In a test of effectiveness of garlic, 49 subjects were treated with doses of raw garlic, and their cholesterol levels were measured before and after treatment. The changes in their levels of cholesterol (in mg/dL) have a mean of 0.4 and a standard deviation of 21.0. Construct a 98% confidence level estimate of the mean net change in cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing cholesterol and why?
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 81 subjects were treated with…
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with…
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 64 subjects were treated with…
A: From the provided information, Sample size (n) = 64 Sample mean (x̅) = 0.9 Sample standard deviation…
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A: The 68-95-99.7% Rule:For a normal distribution, the empirical rule (68-95-99.7% Rule) state that…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 81 subjects were treated with…
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Q: A common claim is that garlic lowers cholesterol levels. In a test of the effectiveness of garlic,…
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…
A: From the provided information, Sample size (n) = 48 Sample mean (x̄) = 2.7 Sample standard deviation…
Q: a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with…
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 81 subjects were treated with…
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Q: n a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with…
A: Obtain the 90% confidence interval estimate of the mean net change in LDL cholesterol after the…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 36 subjects were treated with raw…
A: Here Given Sample Size=n= 36 Sample mean=0.3=x¯ Sample SD=s=1.84
Q: ..... What are the null and alternative hypotheses? OA. Ho: H= 0 mg/dL O B. Ho: u>0 mg/dL H1: µ0…
A: Given data : sample size, n =81 sample mean, x̄ = 0.8 sample standard deviation,s=…
Q: Low‑density lipoprotein, or LDL, is the main source of cholesterol buildup and blockage in the…
A: 68-95-99.7% rule:In normal distribution, the empirical rule of 68-95-99.7% rule states that 68% of…
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A: x1 = 139 n1 = 48 s1 = 33 x2 = 140 n2=43 s2=36
Q: In a test of the effectiveness of garlic for lowering cholesterol, 36 subjects were treated with raw…
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 64 subjects were treated with…
A: The following information is provided: Mean difference xd=0.3 Standard deviation sd=2.36…
Q: the test of the effectiveness of garlic for lowering cholesterol. 49 subjects were treated with…
A: Given Data : Sample Size, n = 49.0 Sample Mean, x̄ = 4.4 standard…
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Q: In a test of the effectiveness of garlic for lowering cholesterol,64 subjects were treated with raw…
A: Given: n = 64 d = 0.7 sd = 1.51 α = 0.01 Formula Used: Test-statistic-t = d-μdsdn
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A: Birth weights are normally distributed with a mean of 7.4 pounds and a standard deviation of 1.1…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with…
A: Sample size, Sample mean, Sample standard deviation, The confidence level is 0.90.
Q: n a test of the effectiveness of garlic for loweringcholesterol, 49 subjects were treated with raw…
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with…
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Q: A recent drug study revealed that a new drug on the market reduces cholesterol levels by an average…
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 64 subjects were treated with raw…
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A: * SOLUTION :- The OPTION C is correct answer. * Explanation :- Given that,
Q: In a test of the effectiveness of garlic for lowering cholesterol, 64 subjects were treated with raw…
A: To find: Test statistics P value Final conclusion
Q: In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with…
A:
Q: In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with…
A: Given: d=before - afterd=0.8sd=15.5α=0.01
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 36 subjects were treated with…
A: We have given thatMean(µ) = 0Sample size (n) = 36Sample mean (x̅) = 0.7Standard deviations (s) =…
Q: for lowering cholesterol,45 subjects were treated with garlic in a processed tablet form.…
A: Given Data : Sample Size, n = 45 Sample Mean, x̄ = 5.1 sample standard…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with…
A: Given Information: Sample size n=49 Mean of the differences d¯=0.3 (sample mean) Standard deviation…
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- In 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 0.6 and a standard deviation of 2.43. 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/dL B. Ho: µ= 0 mg/dL H1: µ#0 mg/dL H1: µ 0 mg/dL O D. Ho: µ= 0 mg/dL H1: µ >0 mg/dL H1:µ<0 mg/dL Determine the test statistic. (Round to two decimal places as needed.) Determine the P-value. (Round to three decimal places as needed.)…Researchers conducted an experiment to test the effects of alcohol. Errors were recorded in a test of visual and motor skills for a treatment group of 22 people who drank ethanol and another group of 22 people given a placebo. The errors for the treatment group have a standard deviation of 2.10, and the errors for the placebo group have a standard deviation of 0.83. Use a 0.05 significance level to test the claim that the treatment group has errors that vary significantly more than the errors of the placebo group. Assume that the two populations are normally distributed. What are the null and alternative hypotheses? OA. Ho: 002 OC. Ho: 0²=0 H₁:00 OB. Ho: 0²=0 H₁: 0>0 OD. Ho: 6²=0 H₁:A study was conducted to assess the effects that occur when children are exposed to cocaine before birth. Children were tested at age 4 for object assembly skill, which was described as "a task requiring visual-spatial skills related to mathematical competence." The 182 children born to cocaine users had a mean of 7.4 and a standard deviation of 2.9. The 181 children not exposed to cocaine had a mean score of 8.3 with a standard deviation of 3.1. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Let population 1 be children born to cocaine users. The test statistic is The P-value is Test the claim in part (a) by using a confidence interval. The ________% confidence interval estimate is ____<μ1−μ2< ____.The heights of English men are normally distributed with a mean of 69.2 inches and a standard deviation of 4 inches. According to the empirical rule, what percentage of English men are Under 57.2 inches tall?Researchers conducted an experiment to test the effects of alcohol. Errors were recorded in a test of visual and motor skills for a treatment group of 22 people who drank ethanol and another group of 22 people given a placebo. The errors for the treatment group have a standard deviation of 2.30, and the errors for the placebo group have a standard deviation of 0.74. Use a 0.05 significance level to test the claim that the treatment group has errors that vary significantly more than the errors of the placebo group. Assume that the two populations are normally distributed. => Identify the test statistic. (Round to two decimal places as needed.) Use technology to identify the P-value. (Round to three decimal places as needed.) What is the conclusion for this hypothesis test? reject/fail to reject?A recent drug study revealed that a new drug on the market reduces cholesterol levels by an average of 20 mg/dL with a standard deviation of 6 mg/dL. You think it is more than that. You examine levels of 20 people on the drug (you have access to their before and after levels). The average reduction was 23 mg/dL. With a significance level of 0.05, is your hunch correct? yes or noIn a test of the effectiveness of garlic for lowering cholesterol, 36subjects 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 0.9and a standard deviation of 2.28. 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. a) What are the null and alternative hypotheses? b) Determine the test statistic. c) Determine the P-value. d) (reject/do not reject). There is (sufficient/not sufficient) evidence to conclude that the mean of the population of changes are (=, not, >, <) to 0.In 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 3.2 and a standard deviation of 17,1. Construct a 99% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. 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/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 23.1. 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.In a test of the effectiveness of garlic for lowering cholesterol, 36subjects 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 0.3 and a standard deviation of 2.39. 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.In a test of the effectiveness of garlic for lowering cholesterol, 50 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 5.8 and a standard deviation of 17.3. Construct a 90% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. WWhat does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol? Click here to view a t 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 p? mg/dLIn a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes(beforeminus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 2.8 and a standard deviation of 16.5 .Construct a 95% 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? What is the confidence interval estimate of the population mean mu? mg/dLless than muless than mg/dL (Round to two decimal places as needed.)SEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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