Researchers are testing a new drug for possible side effects. Using a sample of 28 healthy adults, they randomly allocate a tablet that is either a placebo or contains a preparation with 5mg, 10mg or 15mg of the drug. The subjects take their tablet daily for two months. At the end of this time the researchers record the red cell count (× 1012/L) of each subject. 5 points Suppose the researchers want to generate 95% confidence intervals for all pairwise comparisons of mean red cell count from the four dosage levels. Using the Bonferroni method, the critical t value for each interval is: Solution 3.163 2.875 With 4 groups the number of possible pairwise comparisons is (4×3)/2 = 6. The Bonferroni correction replaces the 95% confidence with 100 - 5/6 = 99.167% confidence. This means we want a two-sided tail probability for the t distribution of 1 - 0.99167 = 0.00833. All we need now are the pooled error degrees of freedom that will be used in the pairwise comparisons. With 28 subjects in 4 groups, the error degrees of freedom are 28-4 = 24. You can then use qt (0.00833/2, df=24) in R to find the critical value, f = 2.875. two tailed Response Continuous Height e.g. Explanatory Categorical Sex Another problem Method Two - sample t-test different groups compared . ( 2 groups ) One way ANOVA F-test (+2 groups) see differences somewhere else Welch or Pooled var equal = TRUE ( higher of , but assure sch is same ) Rank sum test Kruskal - wallis test One -sample t-test Paired t- test Sign test signed rank test Linear regression ( t. test ) Correlation + test ( Spearman's correlation ) Two - way ANOVA the pig example . X2 test one tail Continuous Continuous Continuous Continuons Cat xlat Categorical Categorical Cont X Cat Catergoriaal normaal approx Multiple regression to binomial Continuous Continuous Categorical ( 2 values) Two - sample proportions & test One sample proportion & test X2 feit Logistic regression & test
Researchers are interested in factors involved with low birth weight deliveries at an obstetrics clinic. The data below contains information on 75 births from a sample of deliveries at the clinic. O points + BirthWeight.csv Using logistic regression, the test of association between a low birth weight outcome and maternal age has a P-value of: 0.6529 0.7897 Solution Read the data into R using births = read.csv("BirthWeight.csv", as.is=FALSE). The as. is=FALSE option forces R to convert strings into factors. Use gln( ) to obtain the model estimates: summary(glm(LowWeight ~ Age, data=births, family=binomial)) The Pr(>|z|) column then gives the P-values for a two-sided test for each estimate (against a null hypothesis of 0). Here we want the P-value corresponding to Age. Another problem Researchers are interested in factors involved with low birth weight deliveries at an obstetrics cinic. The data below contains information on 75 births from a sample of deliveries at the clinic. 0 points + BirthWeight.csv Based on this data, a 95%