Question

An outlier's effect. Our bodies have a natural elec- trical field that is known to help wounds heal. Does changing the field strength slow healing? A series of experiments with newts investigated this question. In one experiment, the two hind limbs of 12 newts were assigned at random to either experimental or control groups. This is a matched pairs design. The electrical field in the experimental limbs was reduced to zero by applying a voltage. The control limbs were left alone. Here are the rates at which new cells closed a razor cut in each limb, in micrometers per hour: 20.35 NEWTS Newt Control limb 36 41 39 42 44 39 39 56 33 20 49 30 Experimental 28 31 27 33 33 38 45 25 28 33 47 23 12 3 4 5 678 9 10 11 12 im (a) Why is this a matched pairs design? Explain your answer. (b) Make a stemplot of the differences between limbs of the same newt (control limb minus experimental limb). There is a high outlier (c) A good way to judge the effect of an outlier is to do your analysis twice, once with the out- lier and a second time without it. Carry out two t tests to see if the mean healing rate is signifi- cantly lower in the experimental limbs, with one test including all 12 newts and another omitting the outlier. What are the test statistics and their P-values? Does the outlier have a strong influence on your conclusion?

Answer 1

a) Ans: This is a matched pair design. Because the experiment has only two treatment conditions (control and experiment) and grouped into pairs, based on some blocking variable (contol and experiment). Then, within each pair, subjects are randomly assigned to different treatments.

A matched pairs design is a special case of a randomized block design. It can be used when the experiment has only two treatment conditions; and subjects can be grouped into pairs, based on some blocking variable. Then, within each pair, subjects are randomly assigned to different treatments.

b) Ans:

Stem-and-Leaf Display: Difference Stem-and-leaf of Difference N12 Leaf Unit = 1.0 13 2 -0 6 2 -0 4 0 12 (4) 0 5789 4 1 012

It has atlest one high outlier specially the difference value 31.

c) Ans:

Paired T-Test and Cl: Control, Experimental Paired T for Control - Experimental Control Experimental 12 32.58 7.49 Difference 126.42 10.71 N Mean StDev SE Mean 2.63 2.16 3.09 12 39.00 9.12 95% CI for mean difference : (-0.39, 13.22) T-Test of mean difference 0 (Vs not-0) T-Value - 2.08 P-Value0.062 Without outlier Paired T-Test and Cl: Control, Experimental Paired T for Control - Experimental Control Experimental 11 33.27 7.44 Difference 11 4.18 7.76 N Mean StDev SE Mean 2.33 2.24 2.34 11 37.45 7.74 95% CI for mean difference: (-1.03, 9.39) T-Test of mean difference 0 (vs not 0) : T-Value 1.79 P-Value = 0.104

Comment: From the pair t-test with data with outlier and without outlier we can see that the data with outlier has significance difference with 0.10 level of significance. Whereas, it does not have significance difference at 0.10 level of significance. Hence, we can conclude that outlier affect to the analysis.