A consumer organization collected data on two types of automobile batteries, A and B. Both populations are normally distributed with standard deviations of 1.29 for batteries A and 0.88 for batteries B. The summary statistics for 40 observations of each type yielding average mean of 32.25 hours and 29.81 hours for batteries A and batteries B respectively. Construct 90% confidence interval for difference between means life hours for batteries A and batteries B. 5.
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 45 subjects were treated with…
A: The mean is 4.7 and standard deviation is 17.1.
Q: In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with…
A: Given that Sample size =44 subjects Mean difference=3.4 Sample standard deviation (before-after)…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 45 subjects were treated with…
A: Obtain the 99% 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, 44 subjects were treated with…
A: Given Information: Sample size n=44 Sample mean x¯d=4.2 Sample standard deviation sd=19.6 Confidence…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with…
A: From the provided information, Sample size (n) = 50 Sample mean (x̄) = 2.8 Standard deviation (s) =…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with…
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Q: and after the treatment. The changes (before-after) in their levels of LDL cholesterol (in mg/dL)…
A: Given n=sample size=49, Sd=17.1, mean difference x̄d=2.8 Level of significance ɑ=0.05
Q: In a test of the effectiveness of garlic for lowering cholesterol, 47 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, 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, 43 subjects were treated with…
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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.
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A: Given Data : For first group x̄1 = 163.0 s1 = 9.0 n1 = 8 For second…
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, 48 subjects were treated with…
A: According to the provided information,
Q: na test of the effectiveness of garlic for lowering cholesterol 45 subjects were treated with garlic…
A: From the given information, x¯d=4.6sd=15.3n=45 Confidence level is 90% Significance level is 10%.…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with…
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with…
A: From the provided information, Sample size (n) = 50 Sample mean (x̅) = 5.1 Sample standard deviation…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with…
A: Given n=50 Mean=5.5 Standard deviations=17.9 Alpha=0.01
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: 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: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with…
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with…
A: From the provided information, Sample size (n) = 42 Sample mean (x̄) = 4.4 Sample standard deviation…
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A: The mean change (before - after) is 5.8 SD of change is 19.6 Sample size n = 50 then the degrees of…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with…
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Q: What does the confidence interval suggest about the effectiveness of garlic in reducing LDL…
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Q: Construct a 95% confidence interval estimate of the mean net change in LDL cholesterol after the…
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with…
A: We have given that Mean = 3.3, sample size n = 44 standard deviation s = 18.4 significance level =…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…
A: Suppose defines the true population mean net change in LDL cholesterol after the garlic treatment.
Q: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…
A: Given: Number of subjects or sample size (n) = 48 Mean of the changes d¯=2.8 Standard deviation of…
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: In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with…
A: Given that: Sample size, n=44 Sample mean, x¯=4.4 Standard deviation, s=19.4 Confidence level is…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with…
A: It is given that sample mean is 4.4 and the standard deviation is 19.3.
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A: Given: x¯ 1=47.3x¯2=19.3n1=12n2=13s1=28.3s2= 25.8
Q: wat distribution tal W. Rage 1 of the sta w Rage 2 of the sta fidence interval estim
A: According to the sum, in a test of the effectiveness of garlic for lowering cholesterol, 43 objects…
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, 43 subjects were treated with…
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A: Given : Sample Size, n = 42 Sample Mean, x̄ = 5.7 sample standard…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with…
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- 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 3.6 and a standard deviation of 17.5. 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? 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 u? mg/dL <µ< mg/dL (Round to two decimal places as needed.)Bone mineral density (BMD) is a measure of bone strength. Studies show that BMD declines after age 45. The impact of exercise may increase BMD. A random sample of 59 women between the ages of 41 and 45 with no major health problems were studied. The women were classified into one of two groups based upon their level of exercise activity: walking women and sedentary women. The 39 women who walked regularly had a mean BMD of 5.96 with a standard deviation of 1.22. The 20 women who are sedentary had a mean BMD of 4.41 with a standard deviation of 1.02. Which of the following inference procedures could be used to estimate the difference in the mean BMD for these two types of womenResearchers interested in determining the relative effectiveness of two different drug treatments on people with a chronic illness established two independent test groups. The first group consisted of 12 people with the illness, and the second group consisted of 14 people with the illness. The first group received treatment 1 and had a mean time until remission of 166 days with a standard deviation of 8 days. The second group received treatment 2 and had a mean time until remission of 163 days with a standard deviation of 9 days. Assume that the populations of times until remission for each of the two treatments are normally distributed with equal variance. Construct a 90% confidence interval for the difference −μ1μ2 between the mean number of days before remission after treatment 1 ( μ1 ) and the mean number of days before remission after treatment 2 ( μ2 ). Then find the lower limit and upper limit of the 90% confidence interval. Carry your intermediate…
- In a test of the effectiveness of garlic for lowering cholesterol, 45 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (beforeafter) in their levels of LDL cholesterol (in mg/dL) have a mean of 3.8 and a standard deviation of 19.2. 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?Researchers interested in determining the relative effectiveness of two different drug treatments on people with a chronic illness established two independent test groups. The first group consisted of 15 people with the illness, and the second group consisted of 11 people with the illness. The first group received treatment 1 and had a mean time until remission of 187 days with a standard deviation of 6 days. The second group received treatment 2 and had a mean time until remission of 162 days with a standard deviation of 8 days. Assume that the populations of times until remission for each of the two treatments are normally distributed with equal variance. Construct a 90% confidence interval for the difference H₁-H₂ between the mean number of days before remission after treatment 1 (H₁) and the mean number of days before remission after treatment 2 (H₂). Then find the lower limit and upper limit of the 90% confidence interval. Carry your intermediate computations to at least three…Two groups of individuals were compared with respect to a high carbohydrate low-fat diet (LF) and a high-fat low carbohydrate diet (LC). Mood was assessed using a total mood disturbance score where a lower score is associated with a less negative mood. Assuming that the random samples come from independent normal distributions with a common but unknown variance, construct a 95% confidence interval for the difference between the two population means, LC less LF. Group LC LF 28+18.56 28±8.96 28±22.37 28+22.30 n 12 13 mean 47.3 19.3 standard deviation 28.3 25.8
- Two groups of individuals were compared with respect to a high carbohydrate low- fat diet (LF) and a high-fat low carbohydrate diet (LC). Mood was assessed using a total mood disturbance score where a lower score is associated with a less negative mood. Assuming that the random samples come from independent normal distributions with a common but unknown variance, construct a 95% confidence interval for the difference between the two population means, LC less LF. Group LC LF 28+22.37 28+18.56 28+22.30 28±8.96 n 12 13 mean 47.3 19.3 standard deviation 28.3 25.8In 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 < µIn a test of the effectiveness of garlic for lowering cholesterol, 47 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 3.7 and a standard deviation of 18.2. Construct a 90% 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 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? Question Viewer mg/dL < p< mg/dL (Round to two decimal places as needed.) What does the confidence interval suggest about the effectiveness of the treatment? O A. The confidence interval limits do not…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?A market research group would like to know the satisfaction of subscribers in terms of the quality of service provided by Smart Communications and Globe Telecom. Using a survey, they gathered data from two independent samples. The data below summarizes the average customer satisfaction score (out of 10) towards the quality of services of their mobile providers. The satisfaction score is assumed to be normally distributed with unknown and UNEQUAL population variances. Statistics Smart Communications Globe Telecom Mean 4.20 3.15 Standard Deviation 1.25 2.25 Sample Size 25 20 a) Is there a difference in the average satisfaction scores of subscribers from Smart Communications and Globe Telecom? Make the comparisons at the 10% and 5% levels of significance. b) On average, are subscribers from Smart Communications more satisfied with their service provider than from the subscribers from Globe Telecom? Use the 5% and 1% levels of significance.In a test of the effectiveness of garlic for lowering cholesterol, 47 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 3.7 and a standard deviation of 17.7. 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? Click here to view a t distribution table. Click here to view page 1 of ihe standard norma distributon table Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean µ? Omg/dL < µSEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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