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Chapter 21: Two-Sample Problems Comparing means of two independent samples. The goal of inference is to compare the responses to two treatments or to compare the characteristics of two populations. Test: the claim that subjects treated with Lipitor have a mean cholesterol level that is lower than the mean cholesterol level for subjects given a placebo. Conditions for Inference Comparing Two Means 1. Two SRSs from two distinct populations. 2. The samples are independent. 3. Populations are Normally distributed. 4. The mean and standard deviations of the populations are unknown. Inferences about two means: Independent samples (Independent Samples with σ1 and σ2 unknown and not assumed equal.) Objectives. (1) Test a claim about two independent population means (2) Construct a confidence interval estimate of the difference between two independent population means. Notation. For population j = 1, 2 µj = population mean σj = population standard deviation nj = size of the sample xj = sample mean sj = sample standard deviation Requirements. (1) Both samples are simple random samples. (2) The two samples are independent. (3) Either or both of these conditions is satisfied: The two sample size are both large (n1 > 30 and n2 > 30) or both samples come from populations having normal distributions. (4) σ1 and σ2 are unknown and it is not assumed that σ1 and σ2 are equal. There are two tools: Calculators: Handling formulas. Hypothesis Test Statistic for Two Means : Independent Samples t= (x1 − x2 ) − (µ1 − µ2 ) √ s21 s2 + 2 n1 n2 : Test Statistic where µ1 − µ2 is often assumed to be 0. Confidence Interval Estimate of µ1 − µ2 : Independent Samples (x1 − x2 ) − E < (µ1 − µ2 ) < (x1 − x2 ) + E √ s21 s2 E = t∗ + 2, n1 n2 where df = degree of freedom (for t-distribution) = smaller of (n1 − 1) and (n2 − 1) Computers: Excel, Geogebra, or other computer soft. ◦ Construct “Confidence Interval” or perform “Hypothesis Test” by GeoGebra. 1. Click View ⇒ Probability Calculator. 2. Probability Calculator pops up. 3. Select Statistics on the toolbar. (Distribution / Statistics) 4. Select your desire inferential tool to analyze your data. 5. Enter your value (summary) in the boxes. 6. The result is shown below your summary. Ex. The samples are selected from normally distributed populations. The mean tar content of a simple random sample of 25 unfiltered king size cigarettes is 21.1 mg, with a standard deviation of 3.2 mg. The mean tar content of a simple random sample of 25 filtered 100 mm cigarettes is 13.2 mg with standard deviation of 3.7 mg. Use a 0.05 significance level to test the claim that unfiltered king size cigarettes have a mean tar content greater than that of filtered 100 mm cigarettes. What does the result suggest about the effectiveness of cigarette filters? Four Requirements? Type Sample Size Sample Meas Sample S. D. (mg) ————————————————————————————————– Unfiltered 25 21.1 3.2 Filtered 25 13.2 3.7 Solution. 1. Read English and identify the specific claim or hypothesis. 2. Express the specific claim or hypothesis in symbolic form. H0 : µ1 = µ2 Ha : µ1 > µ2 (µ1 ̸= µ2 ) 3. What is the significance level? α = 0.05 4. Calculate Test Statistics. t= (x1 − x2 ) − (µ1 − µ2 ) √ s21 s2 + 2 n1 n2 : Test Statistic where µ1 − µ2 is often assumed to be 0. n1 = 25 x1 = 21.1 s1 = 3.2 t= n2 = 25 x2 = 13.2 s2 = 3.7 (21.1 − 13.2) − (µ1 − µ2 ) 7.9 √ = = 8.074687922 = 8.075 0.97837 3.22 3.72 + 25 25 5. Convert the test statistics to p-value and compare with the significance level. P-value is less than 0.005 α = 0.05 6. Reject H0 or do not reject, and conclude with your English words. Reject H0 . There is sufficient evidence to support the claim that unfiltered king size cigarettes have a mean tar content greater than that of filtered 100 mm cigarettes. Ex. Refer to the previous example, construct a 90% confidence interval estimate of the difference between the mean tar content of unfiltered king size cigarettes and the mean tar content of filtered 100 mm cigarettes. Confidence Interval Estimate of µ1 − µ2 : Independent Samples (x1 − x2 ) − E < (µ1 − µ2 ) < (x1 − x2 ) + E √ s21 s22 ∗ + , E = t n1 n2 where df = degree of freedom (for t-distribution) = smaller of (n1 − 1) and (n2 − 1) Solution. α = 0.1 E = tα/2 √ df = 24 tα/2 = 1.711 s21 s22 + = 1.711(0.97837) = 1.67399 n1 n2 (x1 − x2 ) − E < (µ1 − µ2 ) < (x1 − x2 ) + E (21.1 − 13.2) − 1.67399 < (µ1 − µ2 ) < (21.1 − 13.2) + 1.67399 6.22601 < (µ1 − µ2 ) < 9.57399 6.226 < (µ1 − µ2 ) < 9.574 ⇒ µ1 ̸= µ2 since 0 is not in the confidence interval. Because the confidence interval limits include only positive values, it appears that unfiltered king size cigarettes have a mean tar content greater than that of filtered 100 mm cigarettes.