1690 HW 7

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PHM1690

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Apr 3, 2024

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Part A 1. Part I a. The hypotheses for the research question can be described as follows: i. H 0 : There is not an evident difference in the average IBS-GIS between the “Open Placebo” group and the “No treatment” group ii. H A : There is an evident difference in the average IBS-GIS between the “Open Placebo” group and the “No treatment” group b. The independent two-sample t-test would be the most appropriate to answer the research question. c. The conditions required to complete this test are: i. The two samples are independent ii. Each sample came from a (nearly) normal distribution with mean µ k and population SD σ k (where k = 1,2) d. The test statistic was calculated to be 3.48 x 1 − x 2 = 5.0 − 3.9 = 1.1 ( 1.5 ) 2 37 = 0.060810811 ( 1.3 ) 2 43 = 0.039302326 t = 1.1 √ 0.100113137 = 1.1 0.316406601 = 3.476539353 e. The degrees of freedom were computed to be 71 and the p-value 0.0009. f. Using the t-distribution function, the p-value was 0.00086591. With rounding, this is the same p-value that was found in (e). g. We have enough evidence to support that the difference in the average IBS-GIS between the “Open Placebo” group and the “No Treatment” group is different from zero. h. The alternative test that could be used to answer the research questions is the Wilcoxon Rank-Sum Test 2. Part II a. The confidence interval for the true mean difference of IBS-GIS between the two groups is 0.252 and 1.948. b. We are 99% confident that the true mean difference of IBS-GIS falls between 0.252 and 1.948. c. The confidence interval does not provide convincing evidence that there is a real (average) IBS-GIS difference between Open Placebo and No Treatment groups because the confidence interval does not contain the null value (0). 3. Part III a. The power of the test for the difference in mean IBS-GIS between the two groups at 0.01 significance level is 0.7939. b. The minimum sample size needed to attain he desired power is 102 (51 samples per group). 4. Diamond Prices a. Based on the graphs and summary statistics, it does appear that the 1 carat diamonds are more expensive on average than the 0.99 carat diamonds.

b. The hypotheses can be defined as follows: i. H 0 : the mean price of 1 carat diamonds are the same as the price of 0.99 carat diamonds. ii. H A : the mean price of 1 carat diamonds are more than the price of 0.99 carat diamonds. c. The most appropriate test to answer the research question above is the independent two-sample t-test. d. t = ( x 1 − x 2 ) − null √ s 1 2 n 1 + s 2 2 n 2 = ( 56.81 − 44.51 ) − 0 √ 16.13 2 23 + 13.32 2 23 = 12.3 √ 11.31 + 7.71 = 2.82 e. The degrees of the freedom were calculated to be 42 and the p-value 0.0036. f. We have enough evidence to support that the average price of 1 carat diamonds are more than the average price of 0.99 carat diamonds. Part B a) H0: there is no difference in ERG levels between dominant and recessive genes of RP. H0: μ dominant = μ recessive HA: there is a difference in ERG levels between dominant and recessive genes of RP. HA: μ dominant ≠μ recessive b) The summary statistics for each group are listed in the table below: Group Mean SD Median IQR N Dominant 0.4594894 0.281343 0.429 0.511 47 Recessive 0.3385 0.210293 0.3025 0.3135 32 c) The p-value is 0.0325. We have enough evidence to support that the difference of true means of ERG levels between dominant and recessive genes of RP is different from 0 at significance level 0.05. d) Assessing Normality for Each Group i. Histogram

The graph for the dominant group shows a very loosely defined normal distribution while the histogram for the recessive group shows a slightly right-skewed distribution. ii. Boxplot Both boxplots show an approximately normal distribution with no extreme outliers. iii. QQ Plot The QQ plots above show a relatively normal distribution. There are a few outliers in each plot, but it can still be approximated as normal. iv. Shapiro-Wilks Test Obs. w v z Prob > z Dominant 47 0.94931 2.271 1.743 0.04069 Recessive 32 0.95273 1.577 0.945 0.17221 The results of the Shapiro-Wilks Test yield a large p-value for the Recessive group, but a small p-value for the dominant group. The data of the dominant group has a p-value that is too small to properly assess normality. This data would need to be transformed in order to establish normality.

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e) i. Using ladder 1/Cubic 1/(erg^3) 69.73 0.000 1/Square 1/(erg^2) 67.52 0.000 Inverse 1/erg 54.87 0.000 1/(Square root) 1/sqrt(erg) 37.34 0.000 Log log(erg) 13.97 0.001 Square root sqrt(erg) 3.61 0.165 Identity erg 9.56 0.008 Square erg^2 5.51 0.064 Cubic erg^3 10.10 0.006 Transformation Formula chi2(2) Prob > chi2 . ladder erg if gene_type == "dominant" 1/Cubic 1/(erg^3) 49.86 0.000 1/Square 1/(erg^2) 49.46 0.000 Inverse 1/erg 43.61 0.000 1/(Square root) 1/sqrt(erg) 30.48 0.000 Log log(erg) 9.74 0.008 Square root sqrt(erg) 0.43 0.807 Identity erg 2.45 0.294 Square erg^2 11.81 0.003 Cubic erg^3 20.21 0.000 Transformation Formula chi2(2) Prob > chi2 . ladder erg if gene_type == "recessive" The results of the latter transformation show large p-values for the square root function of both groups. This is the only transformation that both groups have in common. ii. Using gladder

Using the gladder transformation, we see that the dominant group (left) best demonstrates normality using the square root transformation. This is also the case for the recessive group (right). iii. Using qladder Using the qladder transformation, the data for the dominant group (left) best fits the square root transformation. The recessive group (right) also shows a pretty good fit for the square root transformation. f) The p-value was computed to be 0.0779. We do not have enough evidence to support that the difference of true means of ERG levels between dominant and recessive genes of RP is different from zero at significance level 0.05. g) The p-value that was obtained using the Wilcoxon Rank-Sum test was 0.0741. Our data does not present evidence that the mean ERG levels differ by genetic type. h) T-test with Original Data T-test with Transformed Data Nonparametric Test Pros Cons Pros Cons Pros Cons Easier to interpret Cannot guarantee validity of the test if not normally distributed More efficient than nonparametri c method Difficult to find a proper transformation Does not require normality assumption Less efficient than the original t- test Difficult to interpret the results Easy to use and interpret Robust to outliers Ultimately, I would use the t-test with transformed data because it is more efficient than the nonparametric method. The results of the nonparametric method test are invalid because it is based off the original data and not the transformed data.