End of week four quiz stats 7000

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7000

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Jan 9, 2024

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End of week four quiz stats 7000 Use for Questions 1-6 A researcher wishes to determine which, if any, of four new types of flu medication (A, B, C, and D) is more effective at resolving flu symptoms within 72 hours than the currently recommended flu medication (the control). She enrolls the next 100 eligible children who come to her clinic into her study after obtaining parental consent and randomly assigns them to use one of the four new flu medications or the control medication (five groups, 20 children per group). After 72 hours, she determines whether or not the flu symptoms were resolved for each child. She compares the proportion of children for whom flu symptoms were resolved in each group to the control group, pairwise, using an appropriate statistical test (assuming alpha = 0.05). She only cares if the new flu medication is different than the old one, so in each test, she uses a null hypothesis that the proportion of children for whom flu symptoms were resolved in each group is the same in both flu medications, and a two-sided alternative hypothesis that the proportion of children for whom flu symptoms were resolved in each group is different between the two flu medications. The table below displays the results from each of the tests. Treatment p -value A vs. Control 0.3521 B vs. Control 0.1867 C vs. Control 0.0495 D vs. Control 0.0021 1. Give one potential benefit of using random assignment in this study. a. Random assignment helps to eliminate bias and by randomly assigning participants to different groups we can help ensure that any differences in the outcomes are due to the treatment being tested and not due to pre-existing differences between the groups . 2. Based on the table, which of the comparisons would not contain 0 in the 95% confidence interval for the difference in proportions? Check all that apply. a. Treatment A b. Treatment B c. Treatment C d. Treatment D e. The comparisons between C and Control & D and Control would not have 0 in their confidence intervals for the difference in proportions, because these two comparisons have a p- value that is less than 0.05 (or that is statistically significant). 3. Ignoring the issue of multiple comparisons for the moment, which of the new flu medications, if any, are significantly different than the control medication at resolving flu symptoms within 72 hours? Check all that apply. a. Treatment A b. Treatment B c. Treatment C d. Treatment D e. It appears that the control treatment is different than Treatment C and Treatment D because both p-values are less than 0.05. So, there is evidence that Treatments C and D are significantly more effective than the control medication at resolving flu symptoms within 72 hours
4. Based on the table, which comparison provides the strongest evidence of a difference when compared to the control? a. Treatment A b. Treatment B c. Treatment C d. Treatment d 5. Based on the table, how many comparisons is the researcher making? a. 1 b. 2 c. 3 d. 4 e. 5 6. If the researcher properly corrected for the number of comparisons that were made, which of the new flu medications, if any, are significantly different than the control medication at resolving flu symptoms within 72 hours? Explain your reasoning. a. Treatment D would differ from the control medication in resolving flu symptoms within 72 hours. Using Bonferroni correction divide 0.05/4 = 0.0125. When you look at the p-values in the table only Treatment D has a p-value less than the corrected alpha level . Use for Questions 7-9 The researcher also collects data on the change in temperature of each child, comparing their temperature upon entering the clinic to their temperature 72 hours after taking the medications (change in temperature = temperature after – temperature before). The researcher carries out a test of normality for the change in temperature outcome and obtains a p - value of 0.027. 7. State in words the null hypothesis for this test of normality. a. The change in temperature is normally distributed in this population . b. The change in temperature is not normally distributed in this population. c. The change in temperature is normally distributed in the sample. d. The change in temperature is not sample. 8. What can the researcher conclude from the result of this normality test? Explain your reasoning. a. The researcher can conclude that the temperature is not normally distributed in this population. The p-value obtained is 0.027 which is less than the significance level of 0.05 . 9. If there were evidence that the data are not from a normally distributed population, describe one approach the researcher could use to address the non-normality. a. Data transformation is used to stabilize variance and make the data more closely aligned with the normal distribution or improve the interpretability or appearance of graphs. For Question 10 The researcher took a close look at the data and noticed a few outliers in the data for Treatment B. The researcher is concerned that these outliers could have made the treatment effect for Treatment B less pronounced. The researcher thinks that Treatment B may be effective at resolving flu symptoms within 72 hours, but the outliers present in the data are making the results potentially misleading. 10. If you were the researcher, what would you do about the outliers? Justify your answer. a. I would conduct a sensitivity analysis to understand the impact of the outliers on the results. Outliers can distort the results of a study making a treatment making a treatment appear less effective than it actually is.
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