Chapter 10Robustness

Target Goal: I can interpret the results of inference procedures in a randomized experiment.10.2bh.w: hw: pg. 655: 51, 53, 59, 65, 67 - 70

' to b a bag of M&M's for day 1 chapter 11 or you'll miss all the fun!Don t forget ring

Two-sample t procedures Two-sample t procedures are more robust

than the one sample t methods especially when the distributions are not symmetric.

When the sample sizes are equal, probability values from the t table are accurate even for sizes as small as 5.

If the two population distributions have different shapes, larger samples are needed.

Using the Two-Sample t Procedures:The Normal Condition

Sample size less than 15: Use two-sample t procedures if the data in both samples/groups appear close to Normal (roughly symmetric, single peak, no outliers). If the data are clearly skewed or if outliers are present, do not use t.

• Sample size at least 15: Two-sample t procedures can be used except in the presence of outliers or strong skewness.

• Large samples: The two-sample t procedures can be used even for clearly skewed distributions when both samples/groups are large, roughly n ≥ 30.

More Accurate Levels in the t Procedures

Approximate Distribution of the Two-Sample Statistic

The distribution of the two-sample t statistic is close to the t distribution with degrees of freedom df given by (pg. 637) (for sizes n1 and n2 are 5 or larger).

Your calculator will do this for you and uses the value for two sample t tests.

Note: This value is usually not a whole number.

Ex: DDT Poisoning In a randomized comparative experiment,

researchers compared 6 white rats poisoned with DDT with a control group of 6 un-poisoned rats.

Measuring nerve impulses, the researchers compared the height of the second spike as a percent of the first when a nerve in the rats leg was stimulated.

Output from statistical software showed:

Variable: SpikeGroup N Mean Std Dev Std

ErrorDDT 6 17.60 6.3401 2.5883Control 6 9.4998 1.9500 0.7961Variances T DF Prob >Unequal 2.9912 5.9 0.0247Equal 2.9912 10.0 0.0135What does this data tell us? The difference in the means is large, but the small samplemakes the means highly variable.

Perform a significance test to confirm real effect.

Step 1. Identify the population of interest and the parameter you want to draw a conclusion about. State the null and alternative hypothesis in words and symbols.

We want to compare the mean height μ1 DDT electrical spike to the mean height μ2 control for normal rats.

0 1 2:H 0 1 2 : = 0or H

1 2:aH

No diff. in mean heights of rats in DDT vs. control group.

There is a diff. in mean heights of rats in DDT vs. control group.

Step 2. Choose the appropriate inference procedure. If the conditions are met, we will carry out a

two sample t test for .Enter the data into two lists to examine the data. Use normal probability plots (stat plot) and

check for strong skewness or outliers.

1 2

There is no evidence of outliers or skewness shown in the normal prob plots.Both populations are plausibly normal from the 6 observations.

Random: The data come from a randomized experiment.

Independent: Due to random assignment the two groups of rats can be viewed as independent. Individual observations in each group should also be independent. The overall population of white rats is greater than 60.

Step 3. Do - Carry out the inference procedure. Compute the test statistic and P-value. Use software data for “unequal” variances.

t = 2.9912 , df = 5.9, P-value = 0.0247Step 4: Interpret your results:

The low P-value, p = .0247 provides strong evidence against the Ho at the α = 5% level. We reject Ho in favor of Ha and conclude that the mean size of the second spike is large in rats fed DDT.

Read pg. 638 – 651.

In class FR 2003B #6.