Unit 2 Section 2 Answers

Toxicology and Health Sciences

Bailer and Oris (1993, Env Tox and Chemistry, 12, pp787-91) report data from a study of toxic reproductive response in the aquatic organism Ceriodaphnia dubia to the herbicide nitrofen. A measure of reproductive stress in C. Dubia after exposure to the chemical is offspring counts from exposed females. A decrease in mean response in the exposure group suggests a toxic response to the chemical. Data from 20 independent animals were recorded as total number of offspring for three broods per animal.

Control                27 32 34 33 36 34 33 30 24 31
Nitrofen(160mg/l) 29 29 23 27 30 31 30 26 29 29

Perform a t-test to compare the means of these two groups.
Perform the Wilcoxon rank sum test to these data. Do the two test provide the same results?
What is the 95% confidence interval for the difference between the two means?
If we wanted to know the difference between the average offspring count to within plus or minus 1 with 95% confidence, how many females should we use for each group. Assume the pooled variance estimated for the t-test is the true variance.

Note that these are count data. Counts are typically not very normally distributed. Typically the analysis would be performed on transformed data, either the square root of the counts or the natural logarithm of the counts. Transformation of the response could change the results of the two-sample t-test (try if for fun and see how much). The Wilcoxon rank sum test results would not change. Do you know why?

ANSWERS

1. There are two types of t-test that could be used here, depending on whether you assume the populations have equal variances (pooled variance t-test) or have unequal variances (separate variances t-test). We will do both here.

H0: m(control) = m(nitrofen)

HA: m(control) > m(nitrofen)

Using Excel, you can organize the calculation for both t-tests. A printout of the Excel program can be loaded here (the excel program itself can be downloaded from here). Note that the two sample variances are quite different (12.93 and 5.57 respectively) so there is good indication that maybe the separate variances t-test should be used. From the results of this test we conclude that the null hypothesis should be rejected (T-statistic(separate) > t-critical(separate). The conclusion does not seem to differ if we use the pooled variance t-test, hence maybe the differences in variances is not that important. Note that the conclusion does not change if we use the square root transformed counts as well.

2. The results of the Wilcoxon test are also shown on the spreadsheet output. Note that Excel does not do the right job of ranking observations. You have to go back in and replace tied ranks with the correct values. The test statistic is the sum of the control group, and the critical value is obtained from Table 5 in the book as the T_u value for the one-sided test for a=0.05. In this case, T=136 which is greater than the critical value of 127, hence we reject the null hypothesis and conclude that the control treatment does indeed produce median numbers of offspring that is greater than that for the nitrofen group.

3. The 95% confidence intervals are presented on the spreadsheet page in the green boxes. Note that the interval constructed with the separate variances is slightly wider than the one computed using the pooled variances.

4. The calculation for the sample size needed to compute the 95% confidence interval having expected difference of the mean set at E=1 is given in the spreadsheet output (here). Note that 72 females would have to be examined for each group to reach this level of accuracy. The equation for this solution is given on pge 314 in the book.

Notes:

1.If you did this analysis in Minitab, you need to select the STAT>NONPARAMETRICS>Mann-Whitney ... test to get the Wilcoxon Rank Sum test. This is because if you look at the help on the Mann-Whitney test, Minitab calls it the Mann-Whitney-Wilcoxon Rank Sum test. You can download the minitab worksheet for this problem here.

2. To do these test in SAS you would use the program located here. Note that for the Proc TTEST results, the p-values are for two-sided alternatives, hence we would have to halve them to get the equivalent one-side p-value. Also due to the rounding we did in Excel and the approximating done in SAS, the results are not always exactly the same as what we got with Excel. Regardless the conclusions would remain the same.

3. To do these test in SPSS required some experimentation on my part. You can copy down my worksheet from here and try for yourself. The analysis follows from the ANALYSE>COMPARE MEANS> Independent Samples T-test menu items. In the Independent Samples t-test menu you put response as the Test Variable and you can use either "Treatment" or "Group" as the grouping variable. You need to define the levels of the group variable that define the two groups being tested [ for Treatment type in Control and Nitrofen in the two boxes, for "Group" type in 1 and 2 in the boxes.] This gives you a two independent sample t-test result that is almost identical to what we got in the Excel spreasheet. The confidence intervals do not match - I have no explaination for this. The Excel spreadsheet is computed from the equations in the book. The only difference could be in how accurate the two programs compute the critical values from the t distribution. No easy way to check this. The funny thing is that when you try to do the nonparametic Wilcoxon test [Analyse > Nonparametric Tests > Two independent samples ] you have to use the "Group" coding to define the groups. The "Treatment" coding simply does not show up.

As you can see, you have to learn the specific operations of the package you are using and you have to intellegently experiment to get things to work. None of these packages are simply plug and go!