A clinical trial was conducted to test the effectiveness of a drug used for treating in older subjects. Before treatment with the drug 14 subjects had a mean wake time of 104.0 min. After treatment the 14 subjects had smean wake time of 80.9 min and a standard deviation of 21.4 assume that the 14 sample value appear from a normally distributed population of construct a 90% confidence interval estimate of the mean wake time for a population with drug treatment. What does the result suggest about the mean wake time of 104.0 before the treatment Does the drug appear to be effective?
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Confidence interval: Since the population standard deviation is unknown, t distribution can be used…
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A: From the provided information: Sample size, n=43. Sample mean, x¯=3.4. Sample standard deviation,…
Q: a clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: It is given that μ=101x¯=79.5s=22.8α=1-95%=5% The 95% confidence interval for the mean wake time…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Given data: Sample mean = 101.1min Sample size = 13 Sample standard deviation = 20.8 min Confidence…
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A: Given that Sample size =44 subjects Mean difference=3.4 Sample standard deviation (before-after)…
Q: A clinical trial was conducted to test the effectiveness of a drug used for treating insomnia in…
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Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Given data for the after treatment, The formula of constructing the confidence interval about the…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: It is given that a clinical trial was conducted to test the effectiveness of a drug for treating…
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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, 48 subjects were treated with…
A: We have given that, The given information is -The sample mean change in the LDL cholesterol level…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with…
A: The sample size is 44, the mean is 4.9 and the standard deviation is 15.7.
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A:
Q: In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with…
A: Given Information: Sample size n=43 Sample mean x¯d=3.2 Sample standard deviation sd=17.9 Confidence…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 47 subjects were treated with…
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Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 47 subjects were treated with…
A: Let d denotes the changes (before-after) in the levels of LDL cholesterol (in mg/dL) of the…
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A: Given information Sample size (n) = 19 Sample mean x̅ = 77.7 min Standard deviation (s) = 21.1 min…
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Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Since population standard deviation is unknown, Use t-distribution to find t-critical value. Find…
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A: Given information, Number of subjects n=16 Mean wake time before treatment μ0=101.0 mins Mean wake…
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A: The sample size is 15, sample mean is 95.5, and sample standard deviation is 21.8.
Q: In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with…
A: given data sample size (n) = 43sample mean ( x¯ ) = 4.2sample standard deviatio (s) =17.890% ci for…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A:
Q: In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with…
A: Sample mean = x̅ = 4.5 Sample size = n = 43 Sample S.D = s = 16.8 Standard Error = s/√n =…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
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Q: A clinical trial was conducted to test the effectiveness of a drug used for treating insomnia in…
A: Given data,n=15x=91.8s=42.7α=0.10t-critical value at α=0.10 and df=14 is tc=1.761
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
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Q: In a test of the effectiveness of garlic for lowvering cholesterol, 45 subjects were treated with…
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…
A: We have given that, Sample mean (x̄) = 5.5 , standard deviation (s) = 17.5 and sample size (n)…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: given data with drug treatment n = 24 x¯ = 80.4s = 23.990% ci for mean
Q: Find the confidence interval estimate. min <a<min (Round to two decimal places as needed.)
A:
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A:
Q: clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Sample size : n=25 Before treatment , Sample mean : x¯1=103 After treatment , Sample mean : x¯2=96.1…
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A: The sample size is 21, sample mean is 98.7, and sample standard deviation is 22.9. The degrees of…
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Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: GIVEN DATA n = 18 x¯ = 96.9s = 22.295% ci for μ.
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Given data: Sample size(n) = 25 Mean = 82.1 Standard deviation = 24.6 To construct 95% confidence…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: A confidence interval is a type of estimate computed from the statistics of the observed data. This…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with…
A: Since population standard deviation is unknown, use t-distribution to find t-critical value. Find…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 45 subjects were treated with…
A: Since population standard deviation is unknown, Use t-distribution to find t-critical value. Find…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Solution-: Given: We find, he 95% C.I. estimate of the mean wake time gor a population with the…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Given information Sample size (n) = 21 Sample mean x̅ = 78.5 min Standard deviation (s) = 24.8 min…
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Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Given Mean=97.1 Standard deviations=21.3 Sample size=14
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A: It is given that Mean = 97.7Standard deviation = 44.8Sample size = 14
Q: In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with…
A: Given : Sample Size, n = 42 Sample Mean, x̄ = 5.7 sample standard…
A clinical trial was conducted to test the effectiveness of a drug used for treating in older subjects. Before treatment with the drug 14 subjects had a
After treatment the 14 subjects had smean wake time of 80.9 min and a standard deviation of 21.4 assume that the 14 sample value appear from a
What does the result suggest about the mean wake time of 104.0 before the treatment
Does the drug appear to be effective?
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- A clinical trial was conducted to test the effectiveness of a drug used for treating insomnia in older subjects. Before treatment 15 subjects had a mean wake time of 100.0 min. After treatment, the 15 subjects had a mean wake time of 95.7 min and a standard deviation of 20.4 min. Assume that the 15 sample values appear to be from a normally distributed population and construct a 90% confidence interval estimate of the mean wake times for a population with the drug treatments. What does the result suggest about the mean wake time 100.0 min before the treatment? Does the drug appear to be affective.A 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 102.0 min. After treatment, the 20 subjects had a mean wake time of 96.6 min and a standard deviation of 22.6 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 the result suggest about the mean wake time of 102.0 min before the treatment? Does the drug appear to be effective? Construct the 95% confidence interval estimate of the mean wake time for a population with the treatment. min<μ<minA clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 17 subjects had a mean wake time of 101.0 min. After treatment, the 17 subjects had a mean wake time of 96.8 min and a standard deviation of 21.1 min. Assume that the 17 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 101.0 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<____min (Round to one decimal place as needed.) What does the result suggest about the mean wake time of 101.0 min before the treatment? Does the drug appear to be effective? The confidence interval does not include includes the mean wake time of…
- 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 4.1 and a standard deviation of 19.1. 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, 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/dLA clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 21 subjects had a mean wake time of 100.0 min. After treatment, the 21 subjects had a mean wake time of 96.9 min and a standard deviation of 20.3 min. Assume that the 21 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 100.0 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 < ?? minA 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 101.0 min. After treatment, the 20 subjects had a mean wake time of 91.5 min and a standard deviation of 23.1 min. Assume that the 20 sample values appear to be from a normally distributed population and construct a 99% 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 101.0 min before the treatment? Does the drug appear to be effective? Construct the 99% confidence interval estimate of the mean wake time for a population with the treatment. |min<< '☐ › min (Round to one decimal place as needed.)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 3.9 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 < µIn a test of the effectiveness of garlic for lowering cholesterol,44 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 5.6 and a standard deviation of 17.5. 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? 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.)A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 18 subjects had a mean wake time of 101.0 min. After treatment, the 18 subjects had a mean wake time of 97.7 min and a standard deviation of 21.7min. Assume that the 18 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 101.0 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 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 4.1 and a standard deviation of 19.1. 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 used for treating insomnia in older subjects. After treatment with the drug, 20 subjects had a mean wake time of 92.5 min and a standard deviation of 43.9 min. Assume that the 20 sample values appear to be from a normally distributed population and construct a 90% confidence interval estimate of the standard deviation of the wake times for a population with the drug treatments. Does the result indicate whether the treatment is effective? 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