Solutions, exam in HMET5140 fall 2019
Exercise 1. a) The problem can be solved by doing a chi-square test (2p). Hypotheses:
H0: phigh= pmedium =plow H1: At least one pipj. (2p) Proportion cured in group high: 43/50=0.86
Proportion cured in group medium: 29/45=0.64 Proportion cured in group low: 15/50=0.30
Test statistic: (43−20)
2
20 +(7−30)2
30 +(29−18)2
18 +(16−27)2
27 +(15−20)2
20 +(35−30)2
30 = 56.6 (2p)
Should be compared to 95%-percentile of a chi-square distribution with (3-1)*(2-1)=2 d.f., i.e. 5.99 (2p). Reject H0 since 56.6 is larger than 5.99. (2p) If the sample size was twice as large, the value of the test statistic would be even larger, and one would get an even clearer rejection. This applies to all statistics: Increasing the sample size while having the same effects (proportions in this case) as previously observed, will make it easier to reject the null hypothesis (2p). Numerically in this case, doubling the sample size while maintaining proportions of 0.86, 0.64 and 0.30 will double the value of test statistic calculated above.
b) The test performed in a) only tells you that at least two groups are different. It is relevant to know which groups are different; is it only high vs. low, or is it also high vs.
medium and medium vs. low? This gives useful information about the dose-response relationship (3p). The results show that there is an overall difference in the proportion cured, and this applies to all pairwise comparisons. High vs medium dose are
significantly different with p-value 1%, and medium vs. low is significant with p- value<1% (3p).
c) This is a trend test. The null hypothesis is no monotonuous increasing or decreasing trend in the proportion cured across the dose groups, while the alternative is a trend (2p).
The conclusion is strong evidence for a monotonuous trend, with p-value<1% (2p). The test is relevant, since there is a natural ordering of the exposure variable (dose). Then it is interesting to see if there is a trend or not, as is a slightly more restrictive test than what is done in a) and b) (2p).
Exercise 2. a) The Variance ratio test gives a test on whether the variances in both groups are similar or not, which is one of the assumptions for the standard two-sample t-test (2p). Null hypothesis is that variances are equal. Here it is significant, so this assumption is not fulfilled (2p). This is not surprising since the standard deviations given in the table show a five-fold difference between smokers and non-smokers (2p).
b) Null hypothesis is that the population means in the two groups are equal, and the alternative is that there is a difference (2p). We see that there is a significant difference between smokers and non-smokers in blood pressure, with p-value<0.01 (2p).
c) We have to check if both groups are normally distributed. For non-smokers it does not look too bad. For smokers it looks very skewed, so the assumption is not fulfilled. (2p)
d) This is a non-parametric Mann-Whitney/Wilcoxon rank sum test (1p). The null hypothesis here is that the medians (or, more formally, the distributions) of the two groups are equal, versus the alternative that there is some difference (2p). The conclusion is the same as the t-test, p-value is <0.01 so we reject the null hypothesis (1p).
e) The log-transformation is relevant, as 1) the bloodpressure in smokers is right-skewed, and the log-transformation might reduce that problem (2p) and 2) there was much greater variation in bloodpressure among smokers, and this difference will be reduced by log- transforming the data (2p). On the original scale, the effect of smoking is exp(0.17)=1.18, i.e. 18% higher bloodpressure than non-smokers (2p). The confidence interval is simply (exp(0.03), exp(0.31))=(1.03, 1.36) (2p). The effect of increasing alco consumption by 15 gms/day is exp(15*0.0014)=1.022 or 2.2% higher bloodpressure, however this effect is non-significant (p-value=20.6%) (2p).
f) Both in terms of coefficients on the log-scale and p-values, it is apparent that the model results are very similar (2p). However, the AIC value is 253.4 for the log gamma model, while it was 251.6 for the log-linear model. As lower value indicates better fit of the model, the log-linear seems slightly better (2p). The interpretation of coefficients in the log-gamma is similar to the log-linear model, so the effect of being in age group 61-75 compared to 30-45 (reference group) is exp(0.23)=1.26 or 26% higher bloodpressure (2p).
g) The deviance residuals are supposed to be approximately normal if the gamma GLM model fit is ok. This looks fine in the left plot (2p). The right plot, however, describes variation around the model, which should be free of heteroscedasticity (i.e. be fairly constant for small and large predicted values). The variation seems to be greater for large predicted bloodpressure, indicating that there is still some heteroscedasticity present even when using the gamma GLM model (2p).
Grading:
0-24p=F 25-29p=E 29-35p=D 36-43p=C 44-53p=B 54-60p=A