An experiment is assessing the effect of exercise regimen on cholesterol level decrease. A random sample of 45 individuals is selected, and their cholesterol decrease after following the regimen is recorded. These decreases are assumed to be normally distributed with a known variance of 20. The sample mean decrease is found to be 30 units. Calculate the 99% confidence interval for the average cholesterol decrease of all individuals on this regimen.
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An experiment is assessing the effect of exercise regimen on cholesterol level decrease. A random sample of 45 individuals is selected, and their cholesterol decrease after following the regimen is recorded. These decreases are assumed to be
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- An engineer designed a valve that will regulate water pressure on an automobile engine. The engineer designed the valve such that it would produce a mean pressure of 5.5 pounds/square inch. It is believed that the valve performs above the specifications. The valve was tested on 24 engines and the mean pressure was 5.7 pounds/square inch with a variance of 0.49. A level of significance of 0.01 will be used. Assume the population distribution is approximately normal. Make the decision to reject or fail to reject the null hypothesis. DainThe table below lists weights (carats) and prices (dollars) of randomly selected diamonds. Find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. There is sufficient evidence o support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions. For the prediction interval, use a 95% confidence level with a diamond that weighs 0.8 carats. Weight Price 0.3 $508 a. Find the explained variation. 0.4 $1153 0.5 $1332 Round to the nearest whole number as needed.) . Find the unexplained variation. Round to the nearest whole number as needed.) c. Find the indicated prediction interval.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 womenAn engineer designed a valve that will regulate water pressure on an automobile engine. The engineer designed the valve such that it would produce a mean pressure of 5.5 pounds/square inch. It is believed that the valve performs above the specifications. The valve was tested on 24 engines and the mean pressure was 5.7 pounds/square inch with a variance of 0.49. A level of significance of 0.01 will be used. Assume the population distribution is approximately normal. Make the decision to reject or fail to reject the null hypothesis.Citrus Rental is a popular car rental agency that has a history of having too few cars available, so that its available cars are overdriven. The mean monthly mileage over the years for Citrus cars has been about 1550 miles per month. Recently, though, Citrus purchased thousands of new cars, and the company claims that the average mileage of its cars is now less than in the past. To test this, a random sample of 13 recent mileages of Citrus cars was taken. The mean of these 13 mileages was 1425 miles per month, and the standard deviation was 209 miles per month. Assume that the population of recent monthly mileages of Citrus cars is normally distributed. At the 0.10 level of significance, can it be concluded that the mean recent monthly mileage, μ, of Citrus cars is less than 1550 miles per month? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places and round your answers as specified in the table. (If necessary,…Getting an adequate amount of sleep can improve a variety of physiological functions, and it is known that the average number of hours that people sleep each night is normally distributed. A psychiatrist or a major research center takes an SRS of 25 emotionally disturbed patients and records the average amount of sleep that each gets on a typical night. He summarizes the data and finds that Xbar=6.90 and S= 0.878. What sample size would the researcher need to use in order to decrease the margin of error of a 95% confidence interval to +-0.20 hours?A group of researchers is studying the relationship between cortisol (stress hormone) levels and memory, and they want to see if a sample of 100 adults that has been recruited is a good representation of the population it came from, before they conduct additional research. The population has been found to be normally distributed and have a mean cortisol level of 12 mcg/dL, with a standard deviation of 2 mcg/dL. The sample was found to have a mean cortisol level of 15 mcg/DL, with a standard deviation of 3 mcg/dL. For this assignment, construct a confidence interval to determine if this sample mean is significantly different from the population mean. Explain how you know, based on the confidence interval, and specify the confidence level you used. Be sure to show your work and calculations. This can be tricky with Word, so if necessary you may take a photo of your hand calculations and add it to the Word document.Ritalin has been shown to increase attention span and improve academic performance in children with ADHD (Evans et al., 2001). To demonstrate the effectiveness of the drug, a researcher selects a sample of n = 20 children diagnosed with ADHD and measures each child’s attention span before and after taking the drug. The data show an average increase of attention span of MD = 4.8 minutes with a variance of s2 = 125 for the sample of difference scores. Is this result sufficient to conclude that Ritalin significantly improves attention span? Use a one-tailed test at .05. Compute the 80% confidence interval for the mean chance in attention span for the population.A researcher decides to measure anxiety in group of bullies and a group of bystanders using a 23-item, 3 point anxiety scale. Assume scores on the anxiety scales are normally distributed and the variance among the group of bullies and bystanders are the same. A group of 30 bullies scores an average of 21.5 with a sample standard deviation of 10 on the anxiety scale. A group of 27 bystanders scored an average of 25.8 with a sample standard deviation of 8 on the anxiety scale. You do not have any presupposed assumptions whether bullies or bystanders will be more anxious so you formulate the null and alternative hypothesis based on that.A basketball coach believes that the variance of the heights of adult male basketball players is different from the variance of heights for the general population of men. The sample variance of heights, measured in inches, for a random sample of 19 basketball players is 14.22. The sample variance for a random sample of 18 other men is 34.49. Assume that both population distributions are approximately normal and test the coach’s claim using a 0.10 level of significance. Does the evidence support the coach’s claim? Let male basketball players be Population 1 and men in general be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. H0: σ21=σ22: Ha: σ21⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯σ22 Step 2 of 3: Compute the value of the test statistic. Round your answer to four decimal places. Step 3 of 3: Draw a conclusion and interpret the decision.An HIV clinical trial with 1151 subjects was conducted. We will focus on two variables from this trial: IVDrug and Age. IVDrug is a categorical variable that indicates whether each subject never, previously, or currently uses IV drugs. Age lists the subject's age in year. Thus, we have age for subjects in three treatment groups. Assume that the samples were collected independently and come from a normally distributed population with equal variances. An ANOVA is used to test whether average age in the three IV drug groups is the same. A partially filled-in ANOVA table is listed below. Use a significance level of 0.05. p-value Treatment Error Total df SS MS 76.77 F 7.5586A study was conducted of 90 adult male patients following a new treatment for congestive heart failure. One of the variables measured on the patients was the increase in exercise capacity (in minutes) over a 4-week treatment The previous treatment regime had produced an average increase of μ = 2 minutes. The researchers wanted to evaluate whether the new treatment had increased the value of μ in comparison to the previous treatment. The sample data yielded ?̅ = 2.17 and ? = 1.05: Using α = 05, what conclusions can you draw about the research hypothesis?SEE MORE QUESTIONS
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