Inferencing Between Two Samples: Z Tests & Confidence Intervals for Difference of Means - , Study notes of Data Analysis & Statistical Methods

Download Inferencing Between Two Samples: Z Tests & Confidence Intervals for Difference of Means - and more Study notes Data Analysis & Statistical Methods in PDF only on Docsity! Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2006 Lecture 18: Inferences Based on Two Samples Devore: Section 9.1-9.2 Aug, 2006 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2006 z Tests and Confidence Intervals for a Difference Between Two Population Means • An example of such hypothesis would be µ1 − µ2 = 0 or σ1 > σ2. It may also be appropriate to estimate µ1 − µ2 and compute its 100(1 − α)% confidence interval • Assumptions 1. X1, . . . , Xm is a random sample from a population with mean µ1 and variance σ 2 1 2. Y1, . . . , Yn is a random sample from a population with mean µ2 and variance σ 2 2 3. The X and Y samples are independent of one another Aug, 2006 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2006 The Case of Normal Populations with Known Variances • As before, this assumption is a simplification. • Under this assumption, Z = X̄ − Ȳ − (µ1 − µ2)√ σ21 m + σ22 n (1) has a standard normal distribution • The null hypothesis µ1 − µ2 = 0 is a special case of the more general µ1 − µ2 = ∆0. Replacing µ1 − µ2 in (1) with ∆0 gives us a test statistic. Aug, 2006 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2006 • The following summary considers all possible types of alternatives: 1. Ha : µ1 − µ2 > ∆0 has the rejection region z ≥ zα 2. Ha : µ1 − µ2 < ∆0 has the rejection region z ≤ −zα 3. Ha : µ1 − µ2 6= ∆0 has the rejection region z ≥ zα/2 or z ≤ −zα/2. Aug, 2006 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2006 Example • Consider Ex. 9.1 in Devore. Sample sizes are m = 20 and n = 25. Note that m 6= n...it is not important now but will be later... • Note that the normality suggestion is based on some exploratory data analysis • The hypotheses are H0 : µ1 − µ2 = 0 and Ha : µ1 − µ2 6= 0 • The test statistic is z = x̄ − ȳ√ σ21 m + σ22 n Aug, 2006 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2006 • Similar results can be easily obtained for the other two possible alternatives. In particular, if Ha : µ1 − µ2 < ∆0, we have β(∆ ′ ) = 1 − Φ ( −zα − ∆ ′ − ∆0 σ ) • If µ1 − µ2 6= ∆0, the probability of Type II Error is Φ ( zα/2 − ∆ ′ − ∆0 σ ) − Φ ( −zα/2 − ∆ ′ − ∆0 σ ) Aug, 2006 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2006 Example • Consider Example 9.3 from Devore. Suppose that the probability of detecting a difference .5 between the two means should be .90. Can the .01 level test with m = 20 and n = 25 support this? • For a two-sample test we have β(5) = Φ ( 2.58 − 5 − 0 1.34 ) −Φ ( −2.58 − 5 − 0 1.34 ) = .1251 • Because the rejection region is symmetric, we have β(−5) = β(5), and, therefore, the probability of detecting a difference of .5 is 1 − β(5) = .8749. • We can conclude that slightly larger sample sizes are needed. Aug, 2006 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2006 • To determine a sample size that satisfies P ( Type II Error when µ1 − µ2 = ∆′) = β we need to solve σ21 m + σ22 n = (∆ ′ − ∆0)2 (zα + zβ)2 • For two equal sample sizes this yields m = n = (σ21 + σ 2 2)(zα + zβ) 2 (∆′ − ∆0)2 Aug, 2006 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2006 Example • A company claims that its light bulbs are superior to those of its main competitor. If a study showed that a sample of n1 = 40 of its bulbs has a mean lifetime of 647 hours of continuous use with a standard deviation of 27 hours , while a sample of n2 = 40 bulbs made by its main competitor had a mean lifetime of 638 hours of continuous use with a standard deviation of 31 hours, does this substantiate the claim at the 0.05 level of significance? Aug, 2006 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2006 • H0 : µ1 − µ2 = 0 and Ha : µ1 − µ2 > 0 • Reject H0 if Z > 1.645 • Calculations: z = 647 − 638√ 272 40 + 31 2 40 = 1.38 • Decision: H0 cannot be rejected at α = 0.05; the p-value is 0.0838 Aug, 2006 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2006 Confidence intervals for µ1 − µ2 • Since the test statistic Z that we just described is exactly normal when σ21 and σ 2 2 are known, P  −zα/2 < Z = X̄ − Ȳ − (µ1 − µ2)√ σ21 m + σ22 n < zα/2   = 1−α • The 100(1 − α)% CI is easy to derive from this probability statement; it is x̄ − ȳ ± zα/2σX̄−Ȳ where σX̄−Ȳ is a square root expression. Aug, 2006 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2006 • The point estimate of µB − µA is x̄B − x̄A = 42 − 36 = 6. For α = 0.04, we find the critical value z.02 = 2.05. • Thus, the confidence interval is 6 ± 2.05 √ 64 36 + 36 50 = (3.43, 3.87) Aug, 2006 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2006 The Two-Sample t-test • Assumptions: Both populations are normal, so that X1, . . . , Xm is a random sample from a normal distribution and so is Y1, . . . , Yn. The plausibility of these assumptions can be judged by constructing a normal probability plot of the xis and another of the yis. Aug, 2006 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2006 • When the population distributions are both normal, the standardized variable T = X̄ − Ȳ − (µ1 − µ2)√ S21 m + S22 n has approximately t distribution with ν df • ν can be estimated from data as ν = ( s21 m + s22 n )2 (s21/m) 2 m−1 + (s22/n) 2 n−1 • ν has to be rounded down to the nearest integer...why not up? Aug, 2006 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2006 Example • Consider example 9.6 in Devore. The following table helps to illustrate it: Fabric Type Sample Size Sample Mean Sample Standard Deviation Cotton 10 51.71 .79 Triacetate 10 126.14 3.59 Aug, 2006 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2006 • We assume that porosity distributions for both types of fabric are normal; then, the two-sample t-test(CI) can be used. Note that we do not assume anything about variances of the two populations concerned... • The number of df is ν = ( .6241 10 + 12.881 10 )2 (.6241/10)2 9 + (12.881/10) 2 9 = 9.87 and we use ν = 9 Aug, 2006 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2006 • The resulting CI is 51.71−136.14±(2.262) √ .6241 10 + 12.8881 10 = (−87.06,−81.80) • Conclusion... Aug, 2006 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2006 Remarks • Traditionally, this test has been recommended as the first to use when comparing two different means. It has a number of advantages over the two-sample t test: it is a likelihood ratio test, it is an exact test and it is easier to use! • However, this test has a major problem: it is not robust to the violation of equality of variance assumption. When σ21 = σ 2 2 , its gains in power are small when compared to the two-sample t-test. That is why today it is often recommended to use the two-sample t test in most cases. It is especially true when the sample sizes are different. Aug, 2006 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2006 • It may seem to be a plausible idea that one could first test a hypothesis H0 : σ 2 1 = σ 2 2 and then choose the type of the t test based on the outcome. • Unfortunately, the most common type of test used for this purpose ( we will consider it at the very end of the course) is very sensitive to the violation of normality assumption and often not very reliable as a result. • Yet another warning concerns normality of the data. If the distribution of the data is strongly asymmetric, both of these tests will prove unreliable. The alternative is to use a special class of tests that do not use any distribution assumptions at all (so-called nonparametric tests). Aug, 2006 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2006 Analysis of Paired Data • The data consists of n independently selected pairs (X1, Y1), (X2, Y2),..., (Xn, Yn) with E Xi = µ1 and E Xi = µ2. The differences Di = Xi − Yi are assumed to be normally distributed with mean value µD = µ1 − µ2 and variance σ2D. The last requirement is usually the consequence of X ’s and Y ’s being normally distributed themselves Aug, 2006