Chapter 8 Section 8.5 Testing µ1 - µ2 and p1 - p2 Independent Samples Hypothesis Testing Mr. zboril | Milford PEP

Chapter 8 Sect 8.5 Testing µ1 - µ2 and p1 - p2 Independent Samples You are important to me – don’t ever think otherwise! This Photo by Unknown Author is licensed under CC BY-NC-SA

Chapter 8 Sect 8.5 Testing µ1 - µ2 and p1 - p2 Independent Samples Decisions concerning the value of a parameter will be covered in Chapter 8. Section 8.2 involved testing the mean μ, whereas Section 8.3 addressed testing proportions. Section 8.4 served to enhance the statistical adventure by testing paired differences. Here, we take the means of two populations and test to see if there is a difference. This involved dependent samples. Section 8.5 involves testing dependent samples.

HOW TO TEST µ1 - µ2 WHEN BOTH σ1 AND σ2 ARE KNOWN Requirements Let σ1 and σ2 be the known population standard deviations of populations 1 and 2. Obtain two independent random samples from populations 1 and 2, where x̅1 and x̅2 are sample means from populations 1 and 2 n1 and n2 are sample sizes from populations 1 and 2 If you can assume both population distributions 1 and 2 are normal, any sample sizes n1 and n2 will work. If you cannot assume this, then use sample sizes n1 > 30 and n1 > 30. Procedure In the context of the application, state the null and alternate hypotheses and set the level of significance α. It is customary to use H0: µ1 - µ2 = 0

Interpret your conclusions in the context of the application. HOW TO TEST µ1 - µ2 WHEN BOTH σ1 AND σ2 ARE KNOWN Procedure In the context of the application, state the null and alternate hypotheses and set the level of significance α. It is customary to use H0: µ1 - µ2 = 0 Use µ1 - µ2 = 0 from the null hypothesis together with ̅x1, x̅2, σ1, σ2, n1 and n2 to compute the sample test statistic. 𝑧= (x̅1 − x̅2) − (µ1 − µ2) 𝜎 1 2 𝑛1 + 𝜎 2 2 𝑛2 = (x̅1 − x̅2) 𝜎 1 2 𝑛1 + 𝜎 2 2 𝑛2 Use the standard normal distribution and the type of test, one-tailed or two-tailed, to find the P-value corresponding to the test statistic. Conclude the test. If the P-value < α then reject H0. If the P-value > α then do not reject H0. Interpret your conclusions in the context of the application.

Chapter 8 Sect 8.5 Example 13 A consumer group is testing camp stoves. For each stove, they measure the time to bring 2 quarts of water from 50o to boiling. Two competing models are under consideration. Ten stoves of the first model are being tested and 12 stoves of the second model are tested. The following results were obtained: Model 1: Mean time x̅1 = 11.4 min, σ1 = 2.5 min; n1 = 10. Model 2: Mean time x̅2 = 9.9 min, σ2 = 3.0 min; n2 = 12. Assume that the time required to bring water to a boil is normally distributed for each stove. Is there any difference (either way) between the performances of these two models? Use a 5% level of significance.

Chapter 8 Sect 8.5 Example 13 SOLUTION: Note the level of significance and establish the null and alternate hypotheses. Let µ1 and µ2 be the means of the distribution of times for models 1 and 2, respectively. We set up the null hypothesis to state that there is no difference: H0: μ1 = μ2 or H0: μ1 - μ2 = 0 The alternate hypothesis states that there is a difference: H1: μ1 ≠ μ2 or H0: μ1 - μ2 ≠ 0

Chapter 8 Sect 8.5 Example 13 SOLUTION: Check Requirements What distribution does the sample test statistic follow? The sample test statistic z follows a standard normal distribution because the original distributions from which the samples are drawn are normal. In addition, the population standard deviations of the original distribution are known and the samples are independent.

Chapter 8 Sect 8.5 Example 13 Compute (x̅1 - x̅2) and convert it to a sample test statistic z. We are given the values x̅1 = 11.4 and x̅2 = 9.9. Therefore x̅1 - x̅2 = 11.4 – 9.9 = 1.5. To convert this to a z value, we use σ1 = 2.5 and σ2 = 3.0, n1 = 10, and n2 = 12. From the null hypothesis, µ1 – µ2 = 0 𝑧= (x̅1 − x̅2) − (µ1 − µ2) 𝜎 1 2 𝑛1 + 𝜎 2 2 𝑛2 = 1.5 2.5 2 10 + 3.0 2 12 ≈ 1.28.

Chapter 8 Sect 8.5 Example 13 SOLUTION: Find the P-value of the test statistic and sketch on the standard normal curve. Figure 8-16 and pg. 470 shows the P-value. Since this is a two- tailed test, we use P-value = 2P = 2 x (0.1003) = 0.2006. (discuss Fig. 8-16 on pg. 470) Conclude the test The P-value is 0.2006 > 0.05 for α, we do not reject H0. Interpretation Interpret the results in the context of the problem. At the 5% level of significance, the sample data do not indicate any difference in the population mean times for boiling water for the two stove models.

HOW TO TEST µ1 - µ2 WHEN BOTH σ1 AND σ2 ARE UNKNOWN Requirements Obtain two independent random samples from populations 1 and 2, where x̅1 and x̅2 are sample means from populations 1 and 2 s1 and s2 are sample standard deviations from populations 1 and 2 n1 and n2 are sample sizes from populations 1 and 2 If you can assume both population distributions 1 and 2 are normal, any sample sizes n1 and n2 will work. If you cannot assume this, then use sample sizes n1 > 30 and n2 > 30. Procedure In the context of the application, state the null and alternate hypotheses and set the level of significance α. It is customary to use H0: µ1 - µ2 = 0

HOW TO TEST µ1 - µ2 WHEN BOTH σ1 AND σ2 ARE UNKNOWN Procedure In the context of the application, state the null and alternate hypotheses and set the level of significance α. It is customary to use H0: µ1 - µ2 = 0 Use µ1 - µ2 = 0 from the null hypothesis together with ̅x1, x̅2, s1, s2, n1 and n2 to compute the sample test statistic. 𝑡= (x̅1 − x̅2) − (µ1 − µ2) 𝑠 1 2 𝑛1 + 𝑠 2 2 𝑛2 = (x̅1 − x̅2) 𝑠 1 2 𝑛1 + 𝑠 2 2 𝑛2 The sample test statistic distribution is approximately that of a Student’s t with degrees of freedom d.f. = smaller of n1 – 1 and n2 – 1. Use a Student’s t distribution and the type of test, one-tailed or two-tailed, to find the P- value corresponding to the test statistic. Conclude the test. If the P-value < α then reject H0. If the P-value > α then do not reject H0. Interpret your conclusions in the context of the application.

Chapter 8 Sect 8.5 Example 14 Two competing headache remedies claim to give fast-acting relief. An experiment was performed to compare the mean lengths of time for bodily absorption of brand A and brand B headache remedies. Twelve people were randomly selected and given an oral dosage of brand A. Another twelve people were randomly selected and given an oral dosage of brand B. The length in time in minutes for the drugs to reach a specified level in the blood were recorded. The following results were obtained: Brand A: Mean time x̅1 = 21.8 min, s1 = 8.7 min; n1 = 12. Brand B: Mean time x̅2 = 18.9 min, s2 = 7.5 min; n2 = 12.

Chapter 8 Sect 8.5 Example 14 (cont’d) Brand A: Mean time x̅1 = 21.8 min, s1 = 8.7 min; n1 = 12. Brand B: Mean time x̅2 = 18.9 min, s2 = 7.5 min; n2 = 12. Previous experience with the drug composition allows researchers to assume both distributions are approximately normal. Is there any difference (either way) in the mean time for bodily absorption between the two brands? Use a 5% level of significance. Also, find or estimate the P-value of the sample test statistic.

Chapter 8 Sect 8.5 Example 14 (cont’d) SOLUTION: Note the level of significance and establish the null and alternate hypotheses. Let µ1 and µ2 be the represent the times for Brand A and Brand B, respectively. We set up the null hypothesis to state that there is no difference: H0: μ1 = μ2 or H0: μ1 - μ2 = 0 The alternate hypothesis states that there is a difference: H1: μ1 ≠ μ2 or H0: μ1 - μ2 ≠ 0

Chapter 8 Sect 8.5 Example 14 (cont’d) SOLUTION: Check Requirements What distribution does the sample test statistic follow? We use a Student’s t distribution with degrees of freedom d.f. = smaller of n1 – 1 and n2 – 1. Since both n1 and n2 = 12 d.f. = 12 -1 = 11 The Student’s t is appropriate since the original populations are approximately normal, the population standard deviations are not known, and the samples are independent.

Chapter 8 Sect 8.5 Example 14 (cont’d) Compute the sample test statistic z. We are given the values x̅1 = 21.8 and x̅2 = 18.9. Therefore x̅1 - x̅2 = 21.8 – 18.9 = 2.9. Using s1 = 8.7 and s2 = 7.5, n1 = 12, and n2 = 12. From the null hypothesis, µ1 – µ2 = 0 𝑧= (x̅1 − x̅2) − (µ1 − µ2) 𝑠 1 2 𝑛1 + 𝑠 2 2 𝑛2 = 2.9 8.7 2 12 + 7.5 2 12 ≈ 0.875.

Chapter 8 Sect 8.5 Example 13 SOLUTION: Estimate the P-value and sketch on the standard normal curve. Figure 8-18 on pg. 473 shows the P-value. The d.f = 11. Because the test is a two-tailed test, the P-value is the area to the right of 0.875 and to the left of -0.875. In the Student’s = 2P = 2 x (0.1003) = 0.2006. (discuss Fig. 8-16 on pg. 470) Conclude the test The P-value is 0.2006 > 0.05 for α, we do not reject H0. Interpretation Interpret the results in the context of the problem. At the 5% level of significance, the sample data do not indicate any difference in the population mean times for boiling water for the two stove models.

HOW TO TEST A DIFFERENCE OF PROPORTIONS p1 – p2 Requirements Consider two independent binomial experiments. Binomial Experiment 1 Binomial Experiment 2 n1 = number of trials n2 = number of trials r1 = number of successes r2 = number of successes out of n1 trials out of n2 trials p̂1 = 𝑟1 𝑛1 ; q̂1 = 1 – p̂1 p̂2 = 𝑟2 𝑛2 ; q̂2 = 1 – p̂2 p1 = population probability p2 = population probability of success on a single trial of success on a single trial Use the null hypothesis of no difference, H0: p1 – p2 = 0. In the context of the application, choose the alternate hypothesis. Set the level of significance α. The null hypothesis claims that p1 = p2 therefore pooled best estimates for the population probabilities of success are p̅ = 𝑟1 + 𝑟2 𝑛1 + 𝑛2 and ̅q = 1 - p̅

HOW TO TEST A DIFFERENCE OF PROPORTIONS p1 – p2 (cont’d) Requirements The number of trials should be sufficiently large so that each of the four quantities n1p, n1q, n2p, n2q are > 5. Compute the sample test statistic 𝑧= ̂𝑝1 − ̂𝑝2 𝑝𝑞 𝑛1 + 𝑝𝑞 𝑛2 Use the standard normal distribution and a type of test, one-tailed or two-tailed, to find the P-value corresponding to the test statistic. Conclude the test. If the P-value < α then reject H0. If the P-value > α then do not reject H0. Interpret your conclusions in the context of the application.

Section 8.5 Questions?