Two groups of individuals were comp a high carbohydrate low-fat diet (LF) carbohydrate diet (LC). Mood was ass mood disturbance score where a lowe with a less negative mood. Assuming samples come from independent norr with a common but unknown varianc
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- 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 n 12 13 mean 47.3 19.3 standard deviation 28.3 25.8The respiratory disturbance index (RDI), which is the number of apneic and hypopneic episodes per hour of sleep, was used as a measure of severity of sleep apnea. An RDI of 5 or more indicates sleep-related breathing disturbances. The RDIs for the females and males samples in a study in US are as follows. At the 1% significance level, do the data provide sufficient evidence to conclude that the female patients are less suffer than male from the apnea disease?A researcher notes that there seems to be a difference in the prevalence of high cholesterol among individuals who smoke and consume diets high in processed food and the prevalence of high cholesterol among individuals who do not smoke and do not consume diets high in processed foods. Use the appropriate hypothesis to determine if diet and smoking status are related (independent) at the 5%. This is the first step in determining if confounding has influenced the study’s results. Then interpret your response. Diet Low in Processed Foods Normal Cholesterol High Cholesterol Total Non-smoker 124 55 179 Smoker 69 152 221 Total 193 207 400 Diet High in Processed Foods Normal Cholesterol High Cholesterol Total Non-smoker 64 85 149 Smoker 98 153 251 Total 162 238 400 A) The chi square value of 4.65 is higher than the chi square value of 3.84, therefore we can accept H0 at the 5% level and reject…
- The personnel director of a large corporation wishes to study absenteeism among clerical workers at the corporation’s central office during the year. A random sample of 25 clerical workers reveals a mean absenteeism of 9.7 days with the variance of 16 days. Construct a 95% confidence interval for the average number of absences for clerical workers last year.A group of researchers study the effectiveness of hypnotherapy on duration daily of sleeping hours. Twenty respondents are selected in this study, where ten of them received hypnotherapy treatment and another ten respondents did not receive the treatment. The mean and variance of daily sleeping hours for respondents who received hypnotherapy is 6.9 hours and 1.4333, respectively. The mean and variance of daily sleeping hours for respondents who did not receive hypnotherapy is 5.35 hours and 0.7806, respectively. Assume that the data are normally distributed. (i) Identify the variable used in the study. (ii) By assuming the population variances for both treatments are equal, construct a one-sided lower bound of a 97% confidence interval on the difference in population mean of number of sleeping hours for people who do not receive and receive hypnotherapy treatment. (iii) What is the effect on the width of the confidence interval in (ii) when the sample size increased? State a reason.A simulation model was replicated 100 times yielding a mean of 82.59 with variance of 17.66. Of the 100 replications, 11 replications yielded an outcome over a value of 100. What is the 90% confidence interval on the mean?
- 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 2.7 and a standard deviation of 15.5. 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 < u< mg/dL (Round to two decimal places as needed.)In 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 (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 4.4 and a standard deviation of 19.4. 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 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. St What is the confidence interval estimate of the population mean u? mg/dL «uIn 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 3.2 and a standard deviation of 15.3. 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 μ? mg/dL <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 4.1 and a standard deviation of 17.6. 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.)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 < µA potential investor conducted a 145-day survey in each of the North Mall and South Mall theaters in order to determine the difference between the average daily attendance at these two theaters. The North Mall theater averaged 634 patrons per day; while the South Mall theater averaged 597 patrons per day. From past information, it is known that the variance for North Mall is 1,076; while the variance for the South Mall is 1,329. Develop a 98% confidence interval for the difference between the average daily attendance at the two theaters. (Use North Mall – South Mall. Round your answers to the nearest whole number.) toSEE MORE QUESTIONSRecommended textbooks for youGlencoe Algebra 1, Student Edition, 9780079039897…AlgebraISBN:9780079039897Author:CarterPublisher:McGraw HillGlencoe Algebra 1, Student Edition, 9780079039897…AlgebraISBN:9780079039897Author:CarterPublisher:McGraw Hill