2. Two Sample Problems

3.  Compare the responses of two treatments or compare the characteristics of 2 populations  Separate samples from each population ◦ *Different from Matched pairs  May have 2 different sample sizes  No matching of the units  Example 11.9

4.  Conditions ◦ 1. Two SRS’s; Independent samples (no matching); measuring same variable ◦ 2. Both populations are normally distributed w/ unknown parameters.  Needs ◦ 1. Sample Sizes (n 1, n 2 ) ◦ 2. Sample Statistics (x-bar 1, x-bar 2, s 1, s 2 )  Purpose ◦ Compare 2 means through the test that there is NO difference

5.  The difference of sample means is an unbiased estimator of difference of population means  If 2 population distributions are normal, then the distribution of the sample mean differences is normal. ◦ **Check samples for normality by looking at stem & leaf plot – looking at shape and for outliers

6.  There is a parallel Z statistic that is used when you know the σ of both populations 2 Sample T Statistic Standard Error We have 2 Options for Degrees of Freedom for this situation…

7.  Option 1 ◦ Compute the degrees of freedom based on the data using a formula we’ll examine later  Calculators and Stat Software use this method  Produces df that are not whole #’s (more accurate)  Option 2 ◦ Calculate degrees of freedom for each sample and use the smaller of the two  n 1 - 1  n 2 – 1  Use the smaller of the two…  This is the more conservative of the 2 options and the one we will use initially

8.  Need an SRS from two different normal populations  The Confidence Interval for the Difference of the Two Means is… Using Option 2 to determine degrees of freedom

9.  Researcher’s at Cornell University studied the mating habits of rats with tails versus rats without tails. A random sample of 32 rats with full tails found to father an average of 6.2 rat kiddies with a standard deviation of.62. A random sample of 28 rats with their tails removed found to father an average of 4.8 rats with a standard deviation of.45. Construct a 90% confidence interval for the mean difference between tailed and tailless rat reproduction. 1.1638 – 1.6362

10.  We are testing the H o :µ 1 =µ 2  Write the Hypotheses  Check the Conditions  Calculate the T Statistic  Find the P-Value for the appropriate df  Make statistical decision and interpret results in context

11.  K-Mart claims to pay it’s employees the same as Wal-Mart. The Better Business Bureau randomly sampled 65 K-Mart employees and found an average hourly wage of $8.23 with a standard deviation of $0.85. They sampled 48 Wal-Mart employees and found an hourly wage of $9.12 with a standard deviation of $0.75. Investigate K-Mart’s claim at a 5% significance level. Ho: μ 1 = μ 2 Ha: μ 1 < μ 2 μ 1 = Kmart μ 2 = WalMart t = -5.8898 w/ df = 47 = 40(chart) p <.0005 <  =.05 so we Reject the H o and accept that K-Mart’s average hourly wage is less than Wal-Mart’s.

12.  HW: #39 – 43