Biostatistics Assignment on Malaria Drugs and Hospitalization Cost
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This assignment covers topics on biostatistics including analysis of malaria drugs and hospitalization cost. It includes qnormal plot, confidence interval, independent sample t-test, paired sample t-test and binomial distribution. The assignment is for the course HSH746/HSH946 Biostatistics 1 at a university in a city and state.
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HSH746/HSH946 ASSIGNMENT1 HSH746/HSH946 BIOSTATISTICS 1 Course Professor’s Name University The City and State Date
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HSH746/HSH946 ASSIGNMENT2 Question 1 a.The chart below shows the distribution of the data. 81 01 21 41 6 d u r a t i o n 810121416 Inverse Normal Figure 1: Qnormal plot 051 0 810121416810121416 12 F re q u e n c y duration Graphs by drug Figure 2: Drugs Data Distribution The qnormal plot and the histogram above show proof of normality. The histogram of drug 1 shows a bell shape hence normally distributed and ranges from 12 to 17 days Drug 2 seems to be skewed towards the left-hand side. More data points, in this case, are on the less than 13days.
HSH746/HSH946 ASSIGNMENT3 b.Confidence Interval A confidence interval is the range of values from which one is sure that his/her true values in the dataset will be found. It is given by the formula; Confidence interval (C.I) is given by∓zwhere-population mean, z is the z score and () the standard error (σ √n) Stata commandqnorm duration by drug, sort : summarize duration histogram duration, frequency by(drug) by drug, sort : summarize duration, detail duration2310.05609.18475259.67293410.43924 VariableObsMeanStd.Err.[95%Conf.Interval] ->drug=2 duration2313.92652.211742813.4873914.36565 VariableObsMeanStd.Err.[95%Conf.Interval] ->drug=1 Figure 3: Analysis of Confidence Interval For Drug 1 Mean is the overall average of the sum of malaria symptom duration divided by the number of observation done for each drug (n=23) Mean =∑X N= 320.31/23=13.9 Standard deviation=√(∑(X−Xbar)2 N−1)=√22.69/ (23-1) = 1.0 Since z= (1.96*1.0/√23¿=0.42 (where z score =1.96)
HSH746/HSH946 ASSIGNMENT4 Upper Limit is C.I13.9(0.42) = 14.4 Lower Limit is C.I13.9(0.42) = 13.5 Confidence Interval [13.5, 14.4] For Drug 2 Mean= 231.29/23=10.1 Standard deviation =√(∑(X−Xbar)2 N−1)=√17.27/ (23-1) = 0.9 andz= (1.96*0.9/√23¿=0.36 (where z score =1.96) Upper Limit is C.I10.1(0.36) = 10.4 Lower Limit is C.I10.1(0.36) = 9.7 Confidence interval [9.7, 10.4] c.Drug 1 takes between 13.5 to 14.4 days to reduce symptoms for malaria while drug 2 takes a duration of 9.7 to 10.4 days. This shows that drug 2 has significantly reduced the number of days at which the malaria symptoms were experienced hence can be preferred to drug 1 since its more effective. Question 2 a.In this scenario, the confidence intervals show the range of duration of time a drug will use to reduce the malaria symptoms.According to the finding above has a higher mean and takes a longer duration to reduce the malaria symptoms. Drug two has a smaller mean and the malaria symptoms take a narrower range of time to be fully eradicated. Drug 2 is therefore recommended.
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HSH746/HSH946 ASSIGNMENT5 Question 3 a.An independent sample t-test can be used when the data involved is from independent population and was obtained in a random manner where each observation had an equal chance. Since the two drugs were obtained from independent populations, an independent sample t-test can be carried out to compare the means of the two drugs and provide if the two variables are significantly different B and c TheNull Hypothesis is H0: (μ1- μ2) = 0where μ1is the mean before and μ2 is the mean after the intervention and theAlternative Hypothesis (H1): (μ1- μ2) >0 Stata command used: ttest duration, by(drug) Pr(T<t)=1.0000Pr(|T|>|t|)=0.0000Pr(T>t)=0.0000 Ha:diff<0Ha:diff!=0Ha:diff>0 Ho:diff=0degreesoffreedom=44 diff=mean(1)-mean(2)t=13.7731 diff3.870435.28101343.304094.43678 combined4611.9913.32019862.17169211.3463912.63622 22310.05609.1847525.88604199.67293410.43924 12313.92652.21174281.01548313.4873914.36565 GroupObsMeanStd.Err.Std.Dev.[95%Conf.Interval] Two-samplettestwithequalvariances Figure 4: Independent Sample t-test The figure above shows that drug 1 had the highest duration (M=13.9, SD=1.0) in comparison to drug 2 (M=10.1, S.D=0.9).Also, the value oft (44) =13.77, p<.05.Since the p-value is less than 0.05 which is the threshold, in this case, it shows that the results of the t-test are significant. The null hypothesis can therefore be rejected and conclusion made that the mean duration for drug 1
HSH746/HSH946 ASSIGNMENT6 is different from drug 2.This means duration that the mean duration for drug 2 is slightly smaller than drug 1. Question 4 a.This question utilizes the paired sample t-test knowledge. This is where an experiment is measured before and after an intervention. In this case, the hospitalization cost of patients was measured before and after an intervention.This test, therefore, calculates and compares the means of the hospitalization costs at these two intervals and tries to match if there exist statistically significant difference. TheNull Hypothesis is H0: (μ1- μ2) = 0where μ1is the mean before and μ2 is the mean after the intervention and theAlternative Hypothesis (H1): (μ1- μ2)≠0 Stata command used: ttest before == after . Pr(T<t)=1.0000Pr(|T|>|t|)=0.0000Pr(T>t)=0.0000 Ha:mean(diff)<0Ha:mean(diff)!=0Ha:mean(diff)>0 Ho:mean(diff)=0degreesoffreedom=219 mean(diff)=mean(before-after)t=6.9393 diff22025298.283645.64754073.6918113.2332483.32 after22013302.31053.77615630.0211225.4615379.14 before22038600.573617.51453656.4131470.9845730.17 VariableObsMeanStd.Err.Std.Dev.[95%Conf.Interval] Pairedttest Figure 5: Paired Sample t-test The paired sample t-test revealed that the hospitalization cost before and after are different. The mean difference in cost between the two times is 25,298 AUD. A higher mean of (M=38600 SD=53,656) was realized for the hospitalization cost before and (M=13,302, S.D=15,630) after the intervention. The value oft (219) =6.943.77, p<.05.which was considered at alpha
HSH746/HSH946 ASSIGNMENT7 =0.05.This suggests that there exist a statically significant difference in hospitalization cost before and after the intervention. b.There is, therefore, enough statistical evidence that that the new intervention has resulted in a huge decrease in the hospitalization cost hence should be adopted to save resources. Rejecting the null hypothesis implies that the mean differences between the two cost are not equal to 0 Question 5 The distribution represented here is a binomial distribution. This is where there is a probability of success or failure in an experiment. The population proportion is the fraction of the total number of successful events that were registered in an event divided by the total occurrences.i.e proportion(p hat) = m / n Considering thatmis the number of successes achieved and n is the sample size a.The total number of children that received vaccination are 513 while they that did not are 487. Proportion=513/1000= 0.513= 0.51 (2.dp) b.The confidence interval formula is; Considering (p hat) =0.51 ; C.I =0.51±1.96 √0.51(1−0.51) 1000 C.I =0.51±1.96(0.0158) C.I =0.51±0.031 Confidence interval is[0.48 and 0.54] This implies that the population proportion for thechildren that received the free vaccination was found to lie between interval 48% to 54%.
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