COH 602 Week 2 Quiz (GRADED A) Questions and Answers. VERIFIED

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Question 1 A researcher suspects the mean trough (the lowest dosage of medication required to see clinical improvement of symptoms) level for a medication used to treat arthritis is higher than was ... previously reported in other studies. If previous studies found the mean trough level of the population to be 3.7 micrograms/mL, and the researcher conducts a study among 93 newly diagnosed arthritis patients and finds the mean trough to be 6.1 micrograms/mL with a standard deviation of 2.4 micrograms/mL, for a level of significance of 1%, what should the researcher’s decision rule look like? Reject H0 if z is less than or equal to –2.326. Reject H0 if z is greater than or equal to 2.326 this should be correct but I put one in yellow. Reject H0 if z is less than or equal to –2.576. Reject H0 if z is greater than or equal to 2.5760 This is the two tailed test . The null and alternative hypothesis is , H0 : = 3.7 Ha : 3.7 = 6.1 = 3.7 = 1.2 n = 93 Test statistic = z = ( - ) / / n = (6.1 - 3.7) / 1.2 / 93 = 19.28 Test statistic = 19.28 P(z > 19.28) = 1 - P(z < 19.28) = 1 - 1 = 0 P-value = 0 = 0.01 P-value < Reject the null hypothesis . There is no evidence to the researcher’s hypothesis, for a level of significance of 1%, should not resemble z = 9.645 Question 2 A researcher suspects the mean trough (the lowest dosage of medication required to see clinical improvement of symptoms) level for a medication used to treat arthritis is higher than was previously reported in other studies. If previous studies found the mean trough level of the population to be 3.7 micrograms/mL, and the researcher conducts a study among 93 newly diagnosed arthritis patients and finds the mean trough to be 6.1 micrograms/mL with a standard deviation of 2.4 micrograms/mL, for a level of significance of 1%, what should the researcher’s conclusion be? We have significant evidence at the 1% level to reject H0 in favor of H1 because –9.64 is less than –2.576 and determine the mean trough level for the medication to be higher than 3.7 micrograms/mL. We have significant evidence at the 1% level to reject H0 in favor of H1 because –9.59 is less than –2.326 and determine the mean trough level for the medication to be higher than 3.7 micrograms/mL. We have significant evidence at the 1% level to reject H0 in favor of H1 because 9.64 is greater than 2.326 and determine the mean trough level for the medication to be higher than 3.7 micrograms/mL. We have significant evidence at the 1% level to reject H0 in favor of H1 because 9.59 is greater than 2.576 and determine the mean trough level for the medication to be higher than 3.7 micrograms/mL. Question 3 A recent study indicated that 18% of people free from cardiovascular disease were considered obese. An investigator hypothesizes that a higher proportion of patients with a history of cardiovascular disease will be considered obese. Including 464 patients in the study would ensure that the power of the test is 80% to detect a 5% difference in the proportion with obesity. True False Here test is right tailed. And we have The effect size is For we have and for 80% power we have So the requried sample size is Hence, n = 664 Answer: true Question 4 For samples with fewer individuals, it is a standard practice to use the t score value in place of a z score value when calculating a confidence interval at any level of confidence. True False Question 5 Assume researchers are trying to find out the mean concentration levels of a specific drug in a population of individuals’ blood during a clinical trail. The researchers find that the sample population has a mean concentration of 225 ng/mL with a standard deviation of 47 ng/mL. Calculate a 95% confidence interval for the mean concentration level of the medication for the population’s blood and interpret it. The researchers are 95.0% confident the true mean concentration of medication in the population’s blood is between 147.92 ng/mL and 302.08 ng/mL. The researchers are 95.0% confident the true mean concentration of medication in the population’s blood is between 147.45 ng/mL and 302.55 ng/mL. The researchers are 95.0% confident the true mean concentration of medication in the population’s blood is between 132.88 ng/mL and 317.12 ng/mL. The researchers are 95.0% confident the true mean concentration of medication in the population’s blood is between 147.69 ng/mL and 302.32 ng/mL. Question 6 Assume a researcher wants to compare the mean Aspartate Amino Transferase (AST) levels in two populations, individuals who drink alcohol and individuals who do not drink alcohol. The mean AST levels for the individuals who don’t drink alcohol is 25 with a standard deviation of 10, and 42 individuals were in the sample. The mean AST levels for individuals who drink alcohol is 29 with a standard deviation of 13, and 53 individuals were in the sample. Construct and interpret a 90% confidence interval demonstrating the difference in means for those individuals who drink alcohol when compared to those who do not drink alcohol. The researchers are 95% confident that the true mean difference in AST values between the population of drinkers and population of non-drinkers is between –0.81 and 8.81. The researchers are 95% confident that the true mean difference in AST values between the population of drinkers and population of non-drinkers is between –0.77 and 8.77. The researchers are 95% confident that the true mean difference in AST values between the population of drinkers and population of non-drinkers is between –0.87 and 8.87. The researchers are 95% confident that the true mean difference in AST values between the population of drinkers and population of non-drinkers is between 0.00 and 8.00. Assume a researcher wants to compare the mean Aspartate Amino Transferase (AST) level in two populations. Individual drink alcohol : Sample size = 53 Sample mean = 29 Sample Standard deviation = 13 Individual do not drink alcohol : Sample size = 42 Sample mean = 25 Sample Standard deviation = 10 df = min ( , ) = 42 = 2.019 at 90 % confidence level. The 90% confidence interval demonstrating the difference in mean for Inter pretation : option 2 The researcher are 90% confident that the true mean difference in AST value between the population of drinkers and non- drinkers is between - 0.77 and 8.77. ans-> The researchers are 95% confident that the true mean difference in AST values between the population of drinkers and population of non-drinkers is between –0.81 and 8.81. Question 7 The degrees of freedom in a chi-squared test equal to the number of participants in the test minus 1. True False Question 8 Assume the above results were obtained in a study used to test the accuracy of the rapid diagnostic test for influenza. Calculate the probability of a false negative. .3007 .4778 .5222 .6993 Question 9 A researcher wanted to know the probability that an individual’s diastolic blood pressure would be above 47. Assume the researcher found that the z score corresponding to the 47 diastolic blood pressure reading is –1.96. What is the probability that a random individual’s diastolic blood pressure would be above 47? .025 .250 .750 .975 Question 10 A clinical trial is being conducted in order to determine the efficacy of a new drug that will be used to treat type II diabetes. The efficacy of the medication will be determined by the blood glucose readings obtained from the patients. If the researcher wants a margin of error less than or equal to 5 mg/dL and the standard deviation for blood glucose readings among type II diabetics was previous documented as 15 mg/dL, how many patients should be recruited for each group of individuals in the study assuming a 95% confidence interval will be used to quantify the mean difference in blood glucose scores between the control group and the treatment group? n for the treatment group = 32 and n for the control group = 13 n for the treatment group = 33 and n for the control group = 33 n for the treatment group = 65 and n for the control group = 65 n for the treatment group = 1729 and n for the control group = 1729 for95% CI crtiical Z = 1.960 first sample standard deviation σ1 = 15.000 second sample standard deviation σ2= 15.000 margin of error E = 5 required sample size n=(z/E)2(σ 2+σ 2) = 70 [3] n for the treatment group = 65 and n for the control group = 65 Question 11 A clinical trial is being conducted in order to determine the efficacy of a new drug that will be used to treat Crohn’s disease. The efficacy of the medication will not only be determined by the physical improvement of symptoms but will also be determined by using a blood test to examine the concentration C-reactive protein (an inflammatory marker) in an individual’s blood. If the researcher wants a margin of error for the level of C-reactive protein to be less than or equal to 2.9 mg/dL and the standard deviation for C-reactive protein concentrations among Crohn’s disease patients was previously documented at 10.1 mg/dL, how many patients should be recruited for each group of individuals in the study, assuming a 95% confidence interval will be used to quantify the mean differences between the control group and the treatment group? n for the treatment group = 12 and n for the control group = 12 n for the treatment group = 14 and n for the control group = 14 n for the treatment group = 47 and n for the control group = 47 n for the treatment group = 94 and n for the control group = 94 for95% CI crtiical Z = 1.960 first sample standard deviation σ1 = 10.100 second sample standard deviation σ2= 10.100 margin of error E = 2.9 required sample size n=(z/E)2(σ12+σ22) 94 = n for the treatment group = 94 and n for the control group = 94 Question 12 The positive predictive value of a test is the probability that an individual will test positive for a specific disease given that the individual has the disease in question. True False this is correct answer Course COH602 Biostatistics Test Week Two Exam Time Elapsed 47 minutes out of 1 hour • Question 1 2 out of 2 points – Reject H0 if z is greater than or equal to 2.326 A researcher suspects the mean trough (the lowest dosage of medication required to see clinical improvement of symptoms) level for a medication used to treat arthritis is higher than was previously reported in other studies. If previous studies found the mean trough level of the population to be 3.7 micrograms/mL, and the researcher conducts a study among 93 newly diagnosed arthritis patients and finds the mean trough to be 6.1 micrograms/mL with a standard deviation of 2.4 micrograms/mL, for a level of significance of 1%, what should the researcher’s decision rule look like? • Question 2 2 out of 2 points - We have significant evidence at the 1% level to reject H0 in favor of H1 because 9.64 is greater than 2.326 and determine the mean trough level for the medication to be higher than 3.7 micrograms/mL A researcher suspects the mean trough (the lowest dosage of medication required to see clinical improvement of symptoms) level for a medication used to treat arthritis is higher than was previously reported in other studies. If previous studies found the mean trough level of the population to be 3.7 micrograms/mL, and the researcher conducts a study among 93 newly diagnosed arthritis patients and finds the mean trough to be 6.1 micrograms/mL with a standard deviation of 2.4 micrograms/mL, for a level of significance of 1%, what should the researcher’s conclusion be? • Question 3 1 out of 1 points - true A recent study indicated that 18% of people free from cardiovascular disease were considered obese. An investigator hypothesizes that a higher proportion of patients with a history of cardiovascular disease will be considered obese. Including 464 patients in the study would ensure that the power of the test is 80% to detect a 5% difference in the proportion with obesity. • Question 4 1 out of 1 points - true For samples with fewer individuals, it is a standard practice to use the t score value in place of a z score value when calculating a confidence interval at any level of confidence. • Question 5 2 out of 2 points - The researchers are 95.0% confident the true mean concentration of medication in the population’s blood is between 132.88 ng/mL and 317.12 ng/mL Assume researchers are trying to find out the mean concentration levels of a specific drug in a population of individuals’ blood during a clinical trail. The researchers find that the sample population has a mean concentration of 225 ng/mL with a standard deviation of 47 ng/mL. Calculate a 95% confidence interval for the mean concentration level of the medication for the population’s blood and interpret it. • Question 6 2 out of 2 points - The researchers are 95% confident that the true mean difference in AST values between the population of drinkers and population of non-drinkers is between –0.77 and 8.77. Assume a researcher wants to compare the mean Aspartate Amino Transferase (AST) levels in two populations, individuals who drink alcohol and individuals who do not drink alcohol. The mean AST levels for the individuals who don’t drink alcohol is 25 with a standard deviation of 10, and 42 individuals were in the sample. The mean AST levels for individuals who drink alcohol is 29 with a standard deviation of 13, and 53 individuals were in the sample. Construct and interpret a 90% confidence interval demonstrating the difference in means for those individuals who drink alcohol when compared to those who do not drink alcohol. • Question 7 1 out of 1 points - false The degrees of freedom in a chi-squared test equal to the number of participants in the test minus 1. • Question 8 2 out of 2 points - .4778 Assume the above results were obtained in a study used to test the accuracy of the rapid diagnostic test for influenza. Calculate the probability of a false negative. • Question 9 2 out of 2 points - .975 A researcher wanted to know the probability that an individual’s diastolic blood pressure would be above 47. Assume the researcher found that the z score corresponding to the 47 diastolic blood pressure reading is –1.96. What is the probability that a random individual’s diastolic blood pressure would be above 47? • Question 10 2 out of 2 points - n for the treatment group = 65 and n for the control group = 65 A clinical trial is being conducted in order to determine the efficacy of a new drug that will be used to treat type II diabetes. The efficacy of the medication will be determined by the blood glucose readings obtained from the patients. If the researcher wants a margin of error less than or equal to 5 mg/dL and the standard deviation for blood glucose readings among type II diabetics was previous documented as 15 mg/dL, how many patients should be recruited for each group of individuals in the study assuming a 95% confidence interval will be used to quantify the mean difference in blood glucose scores between the control group and the treatment group? • Question 11 2 out of 2 points - n for the treatment group = 94 and n for the control group = 94 A clinical trial is being conducted in order to determine the efficacy of a new drug that will be used to treat Crohn’s disease. The efficacy of the medication will not only be determined by the physical improvement of symptoms but will also be determined by using a blood test to examine the concentration C-reactive protein (an inflammatory marker) in an individual’s blood. If the researcher wants a margin of error for the level of C-reactive protein to be less than or equal to 2.9 mg/dL and the standard deviation for C-reactive protein concentrations among Crohn’s disease patients was previously documented at 10.1 mg/dL, how many patients should be recruited for each group of individuals in the study, assuming a 95% confidence interval will be used to quantify the mean differences between the control group and the treatment group? • Question 12 1 out of 1 points - false The positive predictive value of a test is the probability that an individual will test positive for a specific disease given that the individual has the disease in question. [Show More]

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