Dentists believe that a diet low in sugary foods can reduce the number of cavities in children. Ten children whose diets are believed to be high in sugar are examined and the mean number of cavities is 3.6 with a standard deviation of 1.6 Twenty children whose diets are believed to be low in sugar are examined and the mean number of cavities is 3.4 with a standard deviation of 0.4 Construct a 99% confidence interval for the true difference between the mean numbers of cavities for children whose diets are high in sugar and those whose diets are low in sugar. Assume that the variances of the two populations are the same. Let Population 1 be children whose diets are believed to be high in sugar and Population 2 be children whose diets are believed to be low in sugar. Round the endpoints of the interval to one decimal place, if necessary.
Q: In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with…
A: The random variable LDL cholesterol level follows normal distribution. We have to construct 99%…
Q: Dentists believe that a diet low in sugary foods can reduce the number of cavities in children.…
A: Obtain the 99% confidence interval for the true difference between the mean numbers of cavities…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…
A: Given information- Sample size, n = 48 Sample mean, x-bar = 3.4 Sample standard deviation, s = 18.9…
Q: A random sample of 21 Camden County College students had a mean age of 24 years, with a standard…
A:
Q: In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with…
A: The changes in the levels of LDL cholesterol have a mean of and a standard deviation of 18.7. The…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with…
A: Given,sample size(n)=42sample mean(x¯)=4.5standard deviation(s)=17.6degrees of…
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, 49 subjects were treated with…
A: Here we need to find a 95% confidence interval for 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:
Q: test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with garlic…
A: Given that Sample size n =43 Sample mean =5.7 Standard deviation =16.1
Q: In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with…
A:
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…
A:
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: Is smoking during pregnancy associated with premature births? To investigate this question…
A: GIven that,According to the bartleby we can strickly answer three subparts
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…
A:
Q: In a test of the effectiveness of garlic for lowering cholesterol, 47 subjects were treated with…
A: The random variable net change in LDL cholesterol follows normal distribution. The sample size is…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…
A: According to the provided information,
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…
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) = 3.5 Sample standard deviation…
Q: nd a standard deviation of 18.9. Construct a 90% confidence interval estimate of the mean net change…
A:
Q: In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with…
A:
Q: a. What is the best point estimate of the population mean net change in LDL cholesterol after the…
A: From the provided information, Sample size (n) = 43 Sample mean (x̄) = 3.7 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: Researchers studying the effects of diet on growth would like to know if a vegetarian diet affects…
A: given data sample size (n) = 9sample mean ( x¯ ) = 42.5sample standard deviation (s) =3.1α = 0.02CL…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…
A: Given that x¯=5.2s=15.6n=48 Confidence level=90% Significance level=1-0.90=0.10 degrees of freedom,…
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, 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: In a test of the effectiveness of garlic for lowvering cholesterol, 45 subjects were treated with…
A:
Q: In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with…
A:
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…
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: n LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the…
A: The mean change (before - after) is 5.8 SD of change is 19.6 Sample size n = 50 then the degrees of…
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: Construct a 95% confidence interval estimate of the mean net change in LDL cholesterol after the…
A:
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, 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, 42 subjects were treated with…
A: In the given, the confidence interval of the before and after use of garlic for lowering cholesterol…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with…
A: It is given that sample mean is 2.7 and standard deviation is 18.9.
Q: In a test of the effectiveness of garlic for lowering cholesterol, 49 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, 49 subjects were treated with…
A: sample size(n)=49Mean()=3.6standard deviation(s)=17.7confidence level=90%
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…
A:
Q: In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with…
A: From the provided information, Sample size (n) = 43 Sample mean (x̄) = 3.1 Sample standard deviation…
Dentists believe that a diet low in sugary foods can reduce the number of cavities in children. Ten children whose diets are believed to be high in sugar are examined and the mean number of cavities is 3.6 with a standard deviation of 1.6 Twenty children whose diets are believed to be low in sugar are examined and the mean number of cavities is 3.4 with a standard deviation of 0.4 Construct a 99% confidence interval for the true difference between the mean numbers of cavities for children whose diets are high in sugar and those whose diets are low in sugar. Assume that the variances of the two populations are the same. Let Population 1 be children whose diets are believed to be high in sugar and Population 2 be children whose diets are believed to be low in sugar. Round the endpoints of the interval to one decimal place, if necessary.
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
- 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 4.7 and a standard deviation of 16.2. 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 µ? Omg/dL <µ< mg/dL (Round to two decimal places 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 5.7 and a standard deviation of 19.6. 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 u? mg/dL < µ< 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 contain 0,…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 (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 5.1 and a standard deviation of 16.8. 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 fabric company is creating a new type of fabric and wants to determine the average strength of this new fabric. They test 20 different strands with a certain number of pounds until the strands tear. They found that the average amount of weight they could hold was 2.7 pounds and the standard deviation is .6 pounds. Find the 95% confidence interval for the mean weight to tear.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, 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 5.5 and a standard deviation of 19.8. 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, 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.)Dentists believe that a diet low in sugary foods can reduce the number of cavities in children. Nineteen children whose diets are believed to be high in sugar are examined and the mean number of cavities is 2.2 with a standard deviation of 0.3. Ten children whose diets are believed to be low in sugar are examined and the mean number of cavities is 3.5 with a standard deviation of 1.4. Construct a 99 % confidence interval for the true difference between the mean numbers of cavities for children whose diets are high in sugar and those whose diets are low in sugar. Assume that the variances of the two populations are the same. Let Population 1 be children whose diets are believed to be high in sugar and Population 2 be children whose diets are believed to be low in sugar. Round the endpoints of the interval to one decimal place, if necessary. Answer 曲 Tables Кeypad Keyboard Shortcuts Lower Endpoint Upper Endpoint =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 2.0 and a standard deviation of 16.5. 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?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 womenIn 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?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? mg/dL < µSEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th…StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C…StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage LearningElementary Statistics: Picturing the World (7th E…StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. FreemanIntroduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. FreemanMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th…StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C…StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage LearningElementary Statistics: Picturing the World (7th E…StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. FreemanIntroduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. Freeman