User:Talgalili/sandbox/Dunnett's test

In statistics, Dunnett's test is a multiple comparison procedure developed by Canadian statistician Charles Dunnett to compare each of a number of treatments with a single control. . Multipule comparisons to a cintrol are also referred to us many to one comparisons.

History
Dunnett's test was developed in 1955; an updated table of critical values was published in 1964.

Multiple Comparisons Problem
The multiple comparisons, multiplicity or multiple testing problem occurs when one considers a set of statistical inferences simultaneously or infers a subset of parameters selected based on the observed values. The major issue in any discussion of multiple-comparison procedures is the question of the probability of Type I errors. Most differences among alternative techniques result from different approaches to the question of how to control these errors. The problem is in part technical; but it is really much more a subjective question of how you want to define the error rate and how large you are willing to let the maximum possible error rate be. Dunnett's test are well known and widely used in multiple comparison procedure for simultaneusly comparing, by interval estimation or hypothesis testing, all active treatments with a control when sampling frome a distribution where the normality assumption is reasonable. Dunnett's test is designed to hold the familywise error rate at or below $$\alpha$$ when performing multiple comparisons of treatment group with control.

Uses of Dunnett’s test
The original work on Multiple Comparisons problem was made by Tukey and Scheffé. Their method was a general one, that considered all kinds of pairwise comparisons. Tukey's and Scheffé's methods allow making any number of comparisons among a set of sample means. But Dunnett's test comes to solve a special case of the general multiple comparisons problem, since we only compare one group with the others - pairwise comparisons of a treatment group with control groups. In the general case, where we compare each of the pairs - we make $$\frac{k(k-1)}{2}$$ comparisons (where k is the number of groups), but in the treatment vs. controls case we will make only $$(k-1)$$ comparisons. If in the case of treatment and control groups we will use the more general Tukey's and Scheffé's methods, it can result in unnecessary wider confidence intervals. Dunnett's test takes under consideration the special structure of comparing treatment against control and result in narrower confidence intervals. it is very common to use Dunnett's test in medical experiments, for example comparing blood count measurements on three groups of animals, one of which served as a control while the other two were treated with two different drugs. Another common use of this method is among agronomists: an agronomist may want to study the effect of certain chemicals added to the soil on crop yield, so he will leave some plots untreated (control plots) and compare them to the plots where chemicals were added to the soil (treatment plots). agronomist tests the effects on crop yield of the addition of chemicals to the soil

Formal description of Dunnett's test
Dunnett's test is performed by computing a Student's t-statistic for each experimental, or treatment, group where the statistic compares the treatment group to a single control group. Since each comparison has the same control in common, the procedure incorporates the dependencies between these comparisons. In particular, the t-statistics are all derived from the same estimate of the error variance which is obtained by pooling the sums of squares for error across all (treatment and control) groups. The formal test statistic for Dunnett's test is either the largest in absolute value of these t-statistics (if a two-tailed test is required), or the most negative or most positive of the t-statistics (if a one-tailed test is required).

In Dunnett's test we can use a common tables of critical values, but more flexible options are nowadays readily available in many statistics packages such as R. The critical values for any given percentage point depend on: whether a one- or- two-tailed test is performed; the number of groups being compared; the overall number of trials.

Assumptions
The analysis considers the case where the results of the experiment are numerical, and the experiment is performed to compare p treatments, with a control group. The results can be summarized as a set of $$(p+1)$$ calculated means of the sets of observations, $$(\bar{X_{0}},...,\bar{X_{p}})$$, while $$(\bar{X_{1}},...,\bar{X_{p}})$$ are referring to the treatment and $$\bar{X_{0}}$$ is refering to the control set of observations, and $$s$$ is an independent estimate of the common standard deviation of all $$p+1$$ sets of observations. All $$\bar{X_{i}}$$ of the $$p+1$$ sets of observations are assumed to be independently and normally distributed with a common variance $$\sigma^2$$ and means $$\mu_{i}$$. There is also an assumption that there is an available estimate $$s^2$$ for $$\sigma^2$$.

Calculation
Calculation Dunett's test's calculation is a procedure that is based on calculating confidence statements about the true or the expected values of the $$p$$ differences $$\bar{X_{i}}-\bar{X_{0}}$$, thus the differences between treatment groups' mean and control group's mean. This procedure enables that the probability of all $$p$$ statements $$\bar{X_{i}}-\bar{X_{0}}$$ being simultaneously correct is equal to a specified value,$$P$$. When calculating one sided upper (or lower) Confidence interval for the true value of the difference between the mean of the treatment and the control group, $$P$$ constitutes the probability that this actual value will be less than the upper (or greater than the lower) limit of that interval. When calculating two-sided confidence interval, $$P$$ constitutes the probability that the true value will be between the upper and the lower limits.

First, we will donate the available N observations by $$X_{ij}$$ when $$i=1...p$$ and $$j=1...N_{i}$$ and estimate the common variance by, for example: $$ s^2= \frac{\sum_{i=0}^{p}\sum_{j=1}^{N_{i}} (X_{ij}-\bar{X_{i}})}{n}$$ when $$\bar{X_{i}}$$ is the mean of group $$i$$ and $$N_{i}$$ is the number of observations in group $$i$$, and $$n=\sum_{i=0}^{p} N_{i}-(p+1)$$ degrees of freedom. As mentioned before, we would like to obtain separate confidence limits for each of the differences $$m_{i}-m_{0}, (i=1...p)$$ such that the probability that all $$p$$ confidence intervals will contain the corresponding $$m_{i}-m_{0}$$ is equal to $$P$$.

We will consider the general case where there are $$p$$ treatment groups and one control group. We will write:

$$z_{i}= \cfrac{\bar{X_{i}}-\bar{X_{0}}-(m_{i}-m_{0})}{\sqrt{ \cfrac{1}{N_{i}}+ \cfrac{1}{N_{0}}}}$$

$$D_{i}= \cfrac{\bar{X_{i}}-\bar{X_{0}}-(m_{i}-m_{0})}{s\sqrt{ \cfrac{1}{N_{i}}+ \cfrac{1}{N_{0}}}}$$

we will also write: $$D_{i}=\frac{z_{i}}{s}$$, which follows the Student's t-statistic distribution with n degrees of freedom. The lower confidence limits with joint confidence coefficient $$P$$ for the $$p$$ treatment effects $$m_{i}-m_{0}, (i=1...p)$$ will be given by:

$$\bar{X_{i}}-\bar{X_{0}}-d_{i}'s\sqrt{ \frac{1}{N_{i}}+\frac{1}{N_{0}}}, i=1...p$$

and the $$p$$ constants $$d_{i}'$$ are chosen so that $$Prob(t_{1}<d_{1}',...,t_{p}<d_{p}')$$ Similarly, the upper limits will be given by:

$$\bar{X_{i}}-\bar{X_{0}}+d_{i}'s\sqrt{ \frac{1}{N_{i}}+\frac{1}{N_{0}}}, i=1...p$$

For bounding $$m_{i}-m_{0}$$ in both directions, the following interval might be taken:

$$\bar{X_{i}}-\bar{X_{0}}\pm d_{i}'s\sqrt{ \frac{1}{N_{i}}+\frac{1}{N_{0}}}, i=1...p$$

when $$d_{i}''$$ are chosen to satisfy $$Prob(|t_{1}|<d_{1}',...,|t_{p}|<d_{p}')$$. The solution to those particular values of $$d_{i}''$$ for two sided test and $$d_{i}'$$ for one sided test is given in the tables. An updated table of critical values was published in 1964.

Breaking strength of fabric
the following example was adapted frome one given by Villars[6]. the data represent measurements on the breaking strength of fabric treated by three different chimical process compared with a standard method of manufacture.

Here, p=3 and N=3. The average variance is $$s^2=19$$, which is an estimate of the common variance of the four sets with (p+1)(N-1)=8 degrees of freedom. It could be directly as foloows:

$$ \frac {55^2+47^2+48^2+55^2+...+41^2-3(50^2+61^2+52^2+45^2)}{8}=\frac{152}{8}=19$$.

The standard deviasion is $$ s=\sqrt{19}=4.36 $$ and the estimated standard error of a difference between two means is $$ s\sqrt{\frac{2}{N}}=4.36\sqrt{\frac{2}{N}}=3.56 $$.

The quantity wich must be added to and/or subtracted from the obsereved differences between the means to give their confedence limits has been called by Tukey an "allwance" and is given by $$A=t_s\sqrt{\frac{2}{N}}$$, where t is obtained from Table 1 if one side limits are desired or from Table 2 if two-sided limits are wanted. For p=3 and d.f.=8, t=2.42 for one side limits and t=2.94 for two-sided limits for p=95%. Amalogous values of t can be detemined from the tables if p=99% confidence is required. For one-sided limits, the allowance is A=(2.42)(3.56)=9 and the experimenter can conclude that: The joint statment consisting of the above three conclusions has a confidence coefficient of 95%, i.e., in the long run, 95% of such joint statments will actually be correct. Upper limits for the three differences could be obtained in an analogous manner. For two-sided limits, the allowance is A=(2.94)(3.56)=11 and the experimenter can conclude that: $$61-50-11=0lbs.$$ and $$61-50-11=22lbs.$$ $$ 52-50-11=-9lbs$$ and $$ 52-50+11=13lbs$$. $$ 45-50-11=-16lbs$$ and $$ 45-50+11=6lbs$$. The joint confidence coefficient for these three statement is greater then 95%. (Due to an approximation made in computing Tables 2a and 2b, the tabulated values of t are somewhat larger then necessary so that the actual p's attained are slightly greater than 95 and 99%.No such approximation was made in computing Tables 1a and 1b).
 * The breaking strength using process 1 exceeds the standard by at least $$61-50-9=2lbs.$$
 * The breaking strength using process 2 exceeds the standard by at least $$52-50-9=-7lbs$$.
 * The breaking strength using process 3 exceeds the standard by at least $$45-50-9=-14lbs$$.
 * The breaking strength using process 1 exceeds the standard by an amount between
 * The breaking strength using process 2 exceeds the standard by an amount between
 * The breaking strength using process 3 exceeds the standard by an amount between