Case 2 Final

Case Study 2: Team 7 Introduction and Discussion of Analysis When we are comparing two or more populations, we need to identify what type of testing we are going to use. For data that is interval, independent samples are collected, the population distribution is normal, and there are one or two factors, we use analysis of variance. We use analysis of variance when we want to analyze the sources of the variation between multiple or single samples. The analysis that we conducted is a one-way analysis of variance or one-way ANOVA. Since there is only one factor, which is the treatments (adenoidectomy, treatment with the medication Sulfafurazole, and a placebo treatment). In this case, the data follows a randomized sampling, which indicates that this follows an ANOVA assumption. This indicates that each group is independent of each other. Even though there are three different categories of treatment or response variables, each treatment is only used once, is independent and does not interact p with the other treatments. Each of the treatments, also known as response variables, are then analyzed between the episodes, visits, prescription, and days. Numerical Statistics Group 1: Surgery Episodes Visits Prescriptions Days Mean 3.3 Mean 2.16666666 7 Mean 3.35 Mean 11.48333 Standard Error 0.170244 Standard Error 0.20469995 5 Standard Error 0.191375 Standard Error 0.615933 Median 3 Median 2 Median 3 Median 11 Mode 2 Mode 2 Mode 2 Mode 11 Standard Deviation 1.318705 Standard Deviation 1.58559903 3 Standard Deviation 1.482382 Standard Deviation 4.770999 Sample Variance 1.738983 Sample Variance 2.51412429 4 Sample Variance 2.197458 Sample Variance 22.76243 Count 60 Count 60 Count 60 Count 60 Numerical statistics for the 60 surgical patients in group 1 show a mean of 3.3 episodes, 2.2 visits, 3.4 prescriptions, and 11.5 days of illness for patients who received surgery to treat otitis media. The standard deviations for these patients are 1.3 for episodes, 1.6 for visits, 1.5 for prescriptions, and 4.8 for days. The variance in episodes is 1.7, for visits is 2.5, for prescriptions is 2.2, and for days is 22.8. This group had the second highest mean for episodes, visits, prescriptions and days, after group 3. Group 1 had the highest standard deviation for

episodes, and the lowest standard deviation for prescriptions. Group 1 has the highest sample variance for episodes, and the lowest sample variance for prescriptions of all three patient groups. Group 2: Rx Sulfafurazole Episodes Visits Prescriptions Days Mean 2.966667 Mean 1.9 Mean 2.96666 7 Mean 11.05 Standard Error 0.140553 Standard Error 0.151639 Standard Error 0.22545 2 Standard Error 0.57313 9 Median 3 Median 2 Median 3 Median 10 Mode 3 Mode 1 Mode 2 Mode 7 Standard Deviation 1.08872 Standard Deviation 1.17459 Standard Deviation 1.74634 4 Standard Deviation 4.43951 8 Sample Variance 1.185311 Sample Variance 1.379661 Sample Variance 3.04971 8 Sample Variance 19.7093 2 Count 60 Count 60 Count 60 Count 60 Numerical statistics for the 60 patients in group 2 that received prescription drug, Sulfafurazole, show a mean of 2.9 episodes, 1.9 visits, 2.9 prescriptions, and 11.1 days of illness for patients who were given Sulfafurazole to treat otitis media. The standard deviation for episodes is 1.1, for visits is 1.2, is 1.7 for prescriptions, and is 4.4 for days. Sample variance for episodes is 1.2, for visits is 1.4, for prescriptions is 3, and for days is 19.7. Group 2 has the lowest means for episodes, visits, prescriptions, and days out of the three groups. Group 2 has the second highest standard deviation for episodes, visits, prescriptions after group 3 and has the lowest standard deviation out of the three groups for days. Similarly, the sample variance in group 2 is second highest of the other groups for episodes. The sample variance for visits in this group is significantly lower than it is for group 1 and group 3. Sample variance for days is the lowest in group 2 as well. Group 3: Placebo Episodes Visits Prescriptions Days Mean 3.45 Mean 2.45 Mean 3.41666 Mean 12.98333 Standard Error 0.167104 Standard Error 0.217101 Standard Error 0.239222 Standard Error 0.685809 Median 3 Median 2 Median 3 Median 12.5 Mode 3 Mode 3 Mode 3 Mode 10 Standard Deviation 1.294382 Standard Deviation 1.681656 Standard Deviation 1.853002 Standard Deviation 5.312255 Sample Variance 1.675424 Sample Variance 2.827966 Sample Variance 3.433616 Sample Variance 28.22006 Count 60 Count 60 Count 60 Count 60 Numerical statistics for the 60 placebo patients, group 3, show a mean of 3.5 episodes, 2.5 visits, 3.4 prescriptions, and 12.9 days spent ill. Standard deviation for episodes is 1.3, for visits is 1.7, for prescriptions is 1.9, and is 5.3 for days. Sample variance for episodes is 1.7, is 2.8

Outliers do not exist in group 2 that are higher or lower for episodes, visits, prescriptions, or days. Episodes have a minimum of 1 and a maximum of 5. Visits have a minimum of 0 and a maximum of 5, showing a slightly wider range than episodes. Prescriptions have a minimum of 0 and a maximum of 7. The prescriptions boxplot also shows that the upper quartile is wider than the lower, with an additional 0.75 of variance than the lower quartile. Days show the widest range between 4 and 22, with the interquartile range (IQR) between 7-14. In group 3, there are several outliers for days. Looking at the mean of 12.98 for days, you can see four outliers that are notably higher. They exist at 24, 26, 27, and 29. There are no lower outliers for episodes, visits, prescriptions, or days. There are no interquartile ranges (IQR’s) for episodes or visits; the numerical data for episodes is 2 and the numerical data for visits is 3 on

the boxplots, creating a more linear visual than a box visual. The IQR for prescriptions is greater, between 2 and 4.75, and the minimums and maximums are between 0-8. The boxplot for days shows a minimum of 5 and a maximum of 20, with the additional outliers previously mentioned. The IQR for days is between 9.25 and 15. Significance Value Interpretation In this case a Type I error indicates that the treatment does not work but looks like it does. The consequence of this error is that parents and doctors will believe the treatment has worked and the child will be sent home with the infection still affecting them. A Type II error will mean that the treatment has worked but looks like it didn’t. The consequence of this error is that the parents and doctors will believe the treatment has not worked and will continue to treat the child even though they are cured. For this case we concluded that the consequences of a Type I error are larger than the consequences of a Type II error. If a Type I error were to occur in this circumstance doctors would be providing a false sense of hope to a distressed parent and child. This error could also bring legal consequences if the family of the child decides to press changes seeing as they were told their child was cured when in reality, they weren’t. Because a Type I error is seen as having a more severe consequence a smaller alpha was chosen to conduct the statistics. The alpha chosen was 0.01 Analysis of Independent Variables Meeting Required Conditions The data for the number of episodes of the illness, number of physician visits, number of prescriptions, and number of days with respiratory infections meets the required conditions for a one-way ANOVA: normal distribution and equal variances. The normality condition was confirmed with histograms as shown below. The histograms are reasonably symmetrical and not severely skewed. The equal variances condition was confirmed using Barlett’s test, the p-value for all the levels were greater than our significance value of 0.01. The null hypothesis was not rejected and therefore there was not enough evidence to conclude that at least 2 of the groups had varying variances.

Analysis for Research Hypothesis and ANOVA’s Episodes A one-way ANOVA test was conducted to determine if there was enough evidence to conclude that the number of episodes differ between the different treatment groups, surgery, drug, and placebo. The hypothesis used for the test was: H o = μ 1 = μ 2 = μ 3 , ∧ H 1 = at least 2 meansdiffer . The alpha used was 0.01 with a rejection rule of: reject if p-value is less than 0.01. The ANOVA table can be seen below. As noted on the ANOVA table the p-value is 0.094. This value is larger than 0.01 and therefore the null hypothesis was not rejected. The conclusion for the number of episodes of the illness is that there was not enough evidence to conclude that the number of episodes differ between the different treatment groups. Visits A one-way ANOVA test provides sufficient evidence to conclude that the number of visits does not differ between the different treatment groups: surgery, drug, and placebo. Like Episodes, the null hypothesis states all three means of the treatment groups are equal while the alternate hypothesis states at least two means differ. This conclusion is made with an alpha of 0.01 and a rejection rule of: reject if p-value is less than 0.01. The ANOVA shown below has a p- value is 0.1349. Given the p-value for visits is larger than 0.01, we do not reject the null hypothesis.

Prescriptions The one-way ANOVA for prescriptions was conducted after determining normal distribution with histograms. With the null hypothesis, we state that the means of treatment groups are equal. With the alternate hypothesis, we state that at least two means differ. Using the alpha 0.01, the p-critical value was determined to be 0.296819. Our rejection rule states that we reject the null hypothesis if the p- values is less than the alpha, 0.01. As noted, our p-value is not less than 0.01 and we then do not reject the null hypothesis. Days Based on the histogram, we can see that the data is normally distributed which satisfies the condition of the use of One-way ANOVA. One-way ANOVA was also conducted in order to identify whether there was enough evidence to conclude the number of days and the different treatments (surgical procedure, the use of drug, and placebo group. The hypothesis test is H 0 : µ 1 = µ 2 = µ 3 and H 1: = at least two means are different. As the data for the ANOVA shows below, the p-critical value is 0.075669 which is greater than 0.01 therefore the null hypothesis was not