Statistics Homework help

Mathematics

Johns Hopkins University

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Statiatics Homework help. For computing the 95% Confidence Inteval, please use 2 instead of 1.96.

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Homework 3, part A 1. (This begins Exercise 1, which consists of Questions 1-5) The following abstract is taken from a September, 2012 article appearing in American Journal of Psychiatry. In this study, the outcome of interest was the change in womens’ responses on the Hamilton Depression Rating Scale between the treatment (antidepressant + oral creatinine supplementation) and control (antidepressant + oral placebo supplementation) Because of randomization, the Hamilton Depression Rating Scale score distributions were nearly identical between the treatment and control groups at baseline (start of the study). As such, the Hamilton scores at the end of the eight-week follow-up period can be compared to assess differences, if any, in the average Hamilton Score change. The Hamilton Depression Rating Scale is an aggregate scale based on user responses to questions about specific psychological and physical symptoms of depression. Higher scores indicate greater severity of depression. A copy of the Hamilton Depression Rating Scale can be found at http://healthnet.umassmed.edu/mhealth/HAMD.pdf Summary statistics about the resulting Hamilton Depression Rating Scale (HDRS) scores at the end of the eight-week clinical trial are as follows: Group n mean standard deviation Treatment 17 5.4 3.0 Control 22 9.8 3.5 - For the “Treatment” group, the estimated standard deviation of the 17 HDRS scores is 3.0. What does this quantity estimate? a. This quantity estimates the variability of individual HDRS scores were the entire population of women with major depressive disorders given the Treatment. b. This quantity estimates the variabililty of sample mean HDRS scores across all possible random samples of n=17 from the population of women with major depressive disorders who were given the Treatment. c. This quantiy estimates how reliable the HDRS scores are in measuring depression. 2. What does the standard error estimate? a. This quantity estimates the variability of individual HDRS scores were the entire population of women with major depressive disorders given the Treatment. b. This quantity estimates the variability in sample mean HDRS scores across all possible random samples of n=17 from the population of women with major depressive disorders if they were given the Treatment. c. This quantiy estimates how likely the sample mean estimate is close to the true population mean. 3. Summary statistics about the resulting Hamilton Depression Rating Scale (HDRS) scores at the end of the eight-week clinical trial are as follows: Group n mean standard deviation Treatment 17 5.4 3.0 Control 22 9.8 3.5 - Interpret the mean difference. a. Subjects in the treatment group had HDRS of 4.4 units greater on average than subjects in the Control group. b. Every subject in the treatment group had a larger HDRS score than every subject in the Control group. c. Subjects in the treatment group had HDRS of 4.4 units less on average than subjects in the Control group. d. Every subject in the treatment group had a lesser HDRS score than every subject in the Control group. 4. Summary statistics about the resulting Hamilton Depression Rating Scale (HDRS) scores at the end of the eight-week clinical trial are as follows: Group n mean standard deviation Treatment 17 5.4 3.0 Control 22 9.8 3.5 - interpret the 95% confidence interval. a. After accounting for sampling variability, the true average HDRS scores were between 2.3 and 6.5 points greater for the populations of women with major depressive disorder had they received the treatment as compared to had they not received the treatment. b. After accounting for sampling variability, the true average HDRS scores were between 2.3 and 6.5 points lower for the populations of women with major depressive disorder had they received the treatment as compared to had they not received the treatment. c. Most of the individuals (about 95%) who were in the treatment population had HDRS scores between -2.3 and -6.5. d. Most of the indiduals (about 95%) who were in the treatment population had HDRS scores between 2.3 and 6.5. 5. Summary statistics about the resulting Hamilton Depression Rating Scale (HDRS) scores at the end of the eight-week clinical trial are as follows: Group n mean standard deviation Treatment 17 5.4 3.0 Control 22 9.8 3.5 - Do the study results show a statistically significant decrease in HDRS scores for the treatment as compared to the control group? a. Yes b. No 6. (This begins exercise 2, which consists of questions 6-10) A 2016 article in JAMA reports the results of a study of treatment outcomes for children with mild gastroenteritis who were given oral rehydration therapy. Enrolled children were randomized to received either rehydration with diluted apple juice (DAJ), or an electrolyte maintenance solution (EMS). As per the study authors: “The primary outcome was a composite of treatment failure defined by any of the following occurring within 7 days of enrollment: intravenous rehydration, hospitalization, subsequent unscheduled physician encounter, protracted symptoms, crossover, and 3%or more weight loss or significant dehydration at inperson follow-up. Secondary outcomes included intravenous rehydration, hospitalization, and frequency of diarrhea and vomiting.” Of the 323 children randomized to DAJ, 54 experienced treatment failure. (17%). Of the 324 children randomized to EMS, 81 experienced treatment failure. (25%) What does the standard error estimate? a. This quantity estimates the variability of individual responses to treatment in the population of children with mild gastroenteritis, were they all treated with DAJ. b. This quantity estimates the potential variability in the sample proportion of children experiencing treatment failure across multiple random samples size n=323 all taken from any population of children. c. This quantity estimates the potential variability in the sample proportion of children experiencing treatment failure across multiple random samples of size n=323 all taken from the population of children with mild gastroenteritis, were they all treated with DAJ. 7. Estimate a 95% CI for the population proportion of children who would experience treatment failure if all children with mild gastroenteritis were treated with DAJ. a. (0.13, 0.21) b. (-0.21, -0.13) c. (0.20, 0.30) d. (-0.30, -0.20) 8. Estimate a 95% CI for the population proportion of children who would experience treatment failure if all children with mild gastroenteritis were treated with EMS. a. (0.13, 0.21) b. (-0.21, -0.13) c. (0.20, 0.30) d. (-0.30, -0.20) 9. Do the 95% confidence intervals from questions 7 and 8 overlap? a. Yes b. No 10. After accounting for sampling variability, is there a statistically significant association between the treatment (DAJ vs. EMS) and treatment failure in the population of children from which the study sample was taken? a. Yes, but only with the relative risk. b. Yes, but only with the risk difference. c. Yes. d. No. 11. (This begins Exercise 3, which consists of questions 11- 17) A 2013 article in the American Journal of Public Health presents the results of study of the relationship between social isolation and mortality. This analysis is based on data collected in the Third National Health and Nutritional Examination Survey (NHANES), and the National Death Index. The following snippet comes from the article abstract: The authors chose to present the results of their analyses separately for females and males. This homework exercise will use only the results for females. The following Kaplan-Meier curve shows the time to death after participation in NHANES 3, when the factors included in computing the social isolation index were measured. The curves are presented separately for the four social index score groups. - Suppose the reference group for computing incidence rate ratios of mortality is the group with index scores of 0. How will the incidence rate ratios for the other 3 social isolation groups relative to this reference group compare to 1 (<, >, or = to)? a. All 3 incidence rate ratios will be greater than 1. b. All 3 incidence rate ratios will be less than 1. c. . Some will be less than 1, and some will be greater than 1. 12. The following portion of an article results table show the hazard ratio (incidence rate ratios) of morality by social isolation index score category, for females. - What is the hazard (incidence rate) ratio of mortality for those with a social isolation score of 0/1 (most isolated) compared to the reference group? a. 1.14 b. 1.29 c. 1.75 d. 1.00 13. Interpret the ratio from item 12 in words. a. Those with a Social network index score of 0-1 (most isolated) have a 75% lower risk of mortality in the follow--up period when compared to those with a Social isolation score of 4 (not isolated) b. Those with a Social network index score of 0-1 (most isolated) have a 75% higher risk of mortality in the follow--up period when compare to those with a Social isolation score of 4 (not isolated) c. Those those with a Social isolation score of 4 (not isolated) have a 75% higher risk of mortality in the follow-up period than Those with a Social network index score of 0-1 (most isolated). d. Those those with a Social isolation score of 4 (not isolated) have a 75% lower risk of mortality in the follow-up period than Those with a Social network index score of 0-1 (most isolated). 14. What is ln (incidence rate ratio) for the ratio from item 12? a. -0.56 b. 0.56 c. -0.24 d. 0.24 15. What is the 95% CI for the incidence rate ratio from item 12? (hint: you do not need to perform an computations to answer this item) a. (0.91, 1.44) b. (1.03, 1.61) c. (1.38, 2.23) d. There is not enough information given to answer this question. 16. Interpret the interval from item 15. a. After accounting for sampling variability the true, population level increase in mortality risk for those who are most isolated compared to those who are not isolated could be between 38% and 123%). b. After accounting for sampling variability the true, population level decrease in mortality risk for those who are most isolated compared to those who are not isolated could be between 38% and 123%) c. After accounting for sampling variability the true, population level increase in mortality risk for those who are not isolated compared to those who are most isolated could be between 38% and 123%). d. After accounting for sampling variability the true, population level decrease in mortality risk for those who are not isolated compared to those who are most isolated could be between 38% and 123%). e. Social isolation is not associated with mortality. 17. What is the general story regarding the social isolation index score and mortality based on these results? a. The estimated incidence (risk) of mortality increases with increasing social isolation, however the increases in risk are not necessarily statistically significantly different between the three lower levels of social isolation. b. The estimated incidence (risk) of mortality decreases with increasing social isolation, however the increases in risk are not necessarily statistically significantly different between the three lower levels of social isolation. c. Increased social isolation is not associated with mortality outcomes in the population from which the sample was taken. Homework 3, Part B 1. The following table summarizes the smoking status and lung cancer status of 1,418 subjects studied in a landmark case-control study done in England in the late 1940’s by British researchers Richard Doll and A. Bradford Hill (This is a classic example of applied biostatistics and epidemiology that has made a huge impact on the medical and public health worlds ) The case-control design is such that the researchers chose subjects on the basis of the outcome of interest (lung cancer) and then assessed the exposure of interest (whether the subject had ever smoked) after the subject was selected. - How many lung cancer (lung carcinoma) were included in this study? 2. - How many control patients were included in this study? 3. The following table summarizes the smoking status and lung cancer status of 1,418 subjects studied in a landmark case-control study done in England in the late 1940’s by British researchers Richard Doll and A. Bradford Hill (This is a classic example of applied biostatistics and epidemiology that has made a huge impact on the medical and public health worlds ) The case-control design is such that the researchers chose subjects on the basis of the outcome of interest (lung cancer) and then assessed the exposure of interest (whether the subject had ever smoked) after the subject was selected. - The researchers chose 1 control for each lung cancer case to create the study sample. Based on this approach, what is the researcher designed risk (prevalence, proportion) of the outcome (lung cancer) in this sample? (please report your answer as a decimal: for example, 0.7 for 70%) 4. Suppose the researchers had instead chosen 3 controls for every lung cancer case: what would the the risk (prevalence, proportion) of the outcome (lung cancer) in the sample be under this study design approach? 5. The following table summarizes the smoking status and lung cancer status of 1,418 subjects studied in a landmark case-control study done in England in the late 1940’s by British researchers Richard Doll and A. Bradford Hill (This is a classic example of applied biostatistics and epidemiology that has made a huge impact on the medical and public health worlds ) The case-control design is such that the researchers chose subjects on the basis of the outcome of interest (lung cancer) and then assessed the exposure of interest (whether the subject had ever smoked) after the subject was selected. - Can the results of this study be used to estimate the risk (prevalence, proportion) of lung cancer in the general population of subjects from which the cases and controls were selected? Why or why not? 6. - What is the overall proportion of smokers (risk of having smoked) among the cases? (Please, report your answer as a decimal: for example, 0.7 for 70%) 7. What is the overall proportion of smokers (risk of having smoked) among the controls? (please report your answer as a decimal: for example, 0.7 for 70%) 8. -Because of the study design, and the fact that the risk of the outcome in the sample is a function of how subjects were selected, and not the risk in the general population being studied, the risks of lung cancer for smokers and non-smokers cannot be estimated for the general population from which the cases and controls were selected. As such, the relative risk of lung cancer for smokers compared to non-smokers cannot be estimated from this case-control study. However, the risks you estimated in sections 6 and 7 (risk of having smoked by case or control status), are legitimate estimates of the risk of having smoked in the general populations of person with and without lung cancer. As such, these can be used to properly estimate the odds ratio of smoking for those with lung cancer relative to those without lung cancer. - What is the estimated odds ratio of smoking for those with lung cancer relative to those without lung cancer? 9. - What is the natural log (ln) of the ratio from question 8? 10. - Estimate the standard error of the ln (odds ratio) from question 9 using the formula provided in the lecture notes. 11. - Estimate a 95% CI for the natural log (ln) of the population level odds ratio. 12. - Estimate a 95% CI for the population level odds ratio. 13. - Does the 95% CI for the odds ratio include the null value for ratios? a. Yes, the 95% CI includes 0. b. Yes, the 95% CI includes 1. c. No the 95% CI does not include 0. d. No the 95% CI does not include 1. 14. - Interpret the odds ratio from question 8 in terms of the odds of lung cancer for smokers compared to non-smokers. 15. - Interpret the 95% confidence interval for odds ratio from question 8 (the interval you reported in question 12) in terms of the odds of lung cancer for smokers compared to nonsmokers. ...
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