1. Chapter 9 Lecture 3 Section: 9.3

2. We will now consider methods for using sample data from two independent samples to test hypotheses made about two population means or to construct confidence interval estimates of the difference between two population means. Definition: Two samples are independent if the sample values selected from one population are not related to, paired or matched with the sample values selected from the other population. If there is some relationship such that each value in one sample is paired with a corresponding value in the other sample, the samples are said to be dependent. Dependent samples are often referred to as matched pairs, or paired samples. For this section, we will only work with independent samples. We will first make our assumptions.

3. Assumptions: 1. The two samples are independent. 2. Both samples are simple random samples. 3. Either or both of these conditions is satisfied: i. The two sample sizes are both large, n 1 > 30 and n 2 > 30 ii. Both samples come from populations having normal distributions. Confidence Intervals Estimate for μ 1 –μ 2 : Independent Samples σ 1 ≠ σ 2 df = smaller of n 1 – 1 or n 2 – 1 “Not Pooled”

4. Hypothesis Testing for 2 Means: Independent Samples σ 1 ≠ σ 2 Test Statistic for H 0 : μ 1 = μ 2 “Not Pooled”

5. As we have discussed before, the population standard deviation σ is rarely known. However, what if there is sufficient evidence to say that the population standard deviations of two population are equal, σ 1 = σ 2. What we can do is pool the two sample variances to obtain an estimate of the common population variance σ 2. The pooled estimate of σ 2 is denoted by s p 2 and is a weighted average s 1 2 and s 2 2. Assumptions: 1. The two populations have the same standard deviation. σ 1 = σ 2 2. The two samples are independent. 3. Both samples are simple random samples. 4. Either or both of these conditions is satisfied: i. The two sample sizes are both large, n 1 > 30 and n 2 > 30 ii. Both samples come from populations having normal distributions.

6. Confidence Intervals Estimate for μ 1 –μ 2 : Independent Samples and σ 1 = σ 2 df = n 1 + n 2 – 2 “Pooled”

7. Hypothesis Testing for 2 Means: Independent Samples and σ 1 = σ 2 Test Statistic for H 0 : μ 1 = μ 2 “Pooled”

8. Examples: 1.A simple random sample of 10 male students results in a sample mean of $5.70 and s = $0.70 on the cost of fast food. Another simple random sample of 12 female students results in a sample mean of $7.90 and s = $0.65 on the cost of fast food. Assume the samples come from normal populations. Construct a 99% confidence interval for the difference of the two population means for the cost of fast food. 2. A sample of 70 subjects is treated with the cholesterol reducing drug. They have an average cholesterol of 110 and a standard deviation of 5. A second and separate group of 81subjects is given a placebo. These subjects have an average cholesterol of 115 with a standard deviation of 7. Construct a 95% confidence interval for the difference of the two population means.

9. 3. A simple random sample of 40 male athletes at Rice University spend an average of 15.00 hours a week studying with s = 1.25. Another simple random sample of 38 female student athletes at Rice University resulted in a sample mean of 19.5 with s = 0.5 in regards to their hours of studying per week. Test the claim that the mean study hours per week of male athletes is less than that of female student athletes at Rice University. Use a 0.05 level of significance.

10. 4. Cost to community hospitals per patient per day are reported by the Hospital Association. Sample statistics from random samples of costs per day form the states of California and Nevada are given below. Test the claim that there is a difference in the mean cost to community hospitals per day in the states of California and Nevada at a 0.01 level of significance.

11. 5. The data below is the measured nicotine contents of randomly selected filtered and unfiltered cigarettes in milligrams. Test the claim that the mean amount of nicotine in filtered cigarettes is equal to the mean amount of nicotine in unfiltered cigarettes. Assume that the data come from normally distributed populations, respectively. Filtered: 1.2 1.3 1.1 1.1 1.0 0.1 1.1 1.0 0.8 1.0 0.9 0.8 1.0 1.0 1.0 0.9 1.2 1.1 0.1 1.2 0.9 Unfiltered: 1.6 1.9 1.6 1.8 1.7 1.7 1.4 1.5

13. Minitab Output: Two-Sample T-Test and CI: Filtered, Nonfiltered Two-sample T for Filtered vs Nonfiltered N Mean StDev SE Mean Filtered 21 0.943 0.309 0.067 Nonfiltered 8 1.650 0.160 0.057 Difference = mu (Filtered) - mu (Nonfiltered) Estimate for difference: -0.707143 95% CI for difference: (-0.944403, -0.469882) T-Test of difference = 0 (vs not =): T-Value = -6.12 P-Value = 0.000 DF = 27 Both use Pooled StDev = 0.2783

14. 6.) Suppose we conduct an experiment to see if two types of music, type- I and type-II, have different effects upon the ability of college students to perform a series of mental tasks requiring concentration. We begin with a fairly homogeneous subject pool of 30 college students, randomly sorting them into two groups. We then have the members of each group, one at a time, perform a series of 40 mental tasks while one or the other types of music is playing in the background. For the members of group A it is music of type-I, while for those of group B it is music of type-II. The following table shows how many of the 40 components of the series each subject was able to complete. Test the claim that the group A has a higher mean than group B. Use α = 0.025 level of significance. Group AGroup B 30 25 26 30 23 26 30 29 28 25 27 27 22 33 22 27 25 24 24 33 24 20 24 30 25 29 21 22 23 26