Independent Samples: Comparing Means Lecture 34 Section 11.4 Wed, Nov 15, 2006

Independent Samples In a paired study, two observations are made on each subject, producing one sample of bivariate data. Or we could think of it as two samples of paired data. Often these are “before” and “after” observations. By comparing the “before” mean to the “after” mean, we can determine whether the intervening treatment had an effect.

Independent Samples On the other hand, with independent samples, there is no logical way to “pair” the data. One sample might be from a population of males and the other from a population of females. Or one might be the treatment group and the other the control group. The samples could be of different sizes.

Independent Samples We wish to compare population means 1 and 2. We do so by comparing sample meansx1 andx2. More specifically, we will usex1 –x2 as an estimator of 1 – 2. If we want to know whether 1 = 2, we test to see whether 1 – 2 = 0 by computingx1 –x2.

The Distributions ofx1 andx2 Let n1 and n2 be the sample sizes. If the samples are large, thenx1 andx2 have (approx.) normal distributions. However, if either sample is small, then we will need an additional assumption. The populations are normal.

Further Assumption We will also assume that the two populations have the same standard deviation. Call it . If this assumption is not supported by the evidence, then it should not be made. If this assumption is not made, then the formulas become much more complicated. See p. 658.

The Distribution ofx1 –x2 Ifx1 andx2 have normal distributions with means 1 and 2 and standard deviations 1/n1 and 2/n2, then x1 –x2 is a normal random variable with the following properties: The mean is 1 – 2. The variance is

The Distribution ofx1 –x2 If we assume that 1 = 2, (call it ), then the standard deviation may be simplified to That is,

The Distribution ofx1 1

The Distribution ofx2 2 1

The Distribution ofx1 –x2 2 1 1 – 2

The Distribution ofx1 –x2 If then it follows that

Example Do Example 11.4, page 699, but Assume that the same sizes are 100, not 10. Then work the same example using the TI-83 and the 2-SampZTest function.

The t Distribution Let s1 and s2 be the sample standard deviations. Whenever we use s1 and s2 instead of , then we will have to use the t distribution instead of the standard normal distribution, unless the sample sizes are large.

Estimating  Individually, s1 and s2 estimate . However, we can get a better estimate than either one if we “pool” them together. The pooled estimate is

x1 –x2 and the t Distribution If we use sp instead of , and the sample sizes are small, then we should use t instead of Z. The number of degrees of freedom is df = df1 + df2 = n1 + n2 – 2. That is

Hypothesis Testing See Example 11.4, p. 699 – Comparing Two Headache Treatments. State the hypotheses. H0: 1 = 2 H1: 1 > 2 State the level of significance.  = 0.05.

The t Statistic Compute the value of the test statistic. The test statistic is with df = n1 + n2 – 2.

Computations

Hypothesis Testing Calculate the p-value. The number of degrees of freedom is df = df1 + df2 = 18. p-value = P(t > 1.416) = tcdf(1.416, E99, 18) = 0.0869.

Hypothesis Testing State the decision. State the conclusion. Accept H0. State the conclusion. At the 5% level of significance, the data do not support the claim that Treatment 1 is more effective than Treatment 2.

The TI-83 and Means of Independent Samples Enter the data from the first sample into L1. Enter the data from the second sample into L2. Press STAT > TESTS. Choose either 2-SampZTest or 2-SampTTest. Choose Data or Stats. Provide the information that is called for. 2-SampTTest will ask whether to use a pooled estimate of . Answer “yes.”

The TI-83 and Means of Independent Samples Select Calculate and press ENTER. The display shows, among other things, the value of the test statistic and the p-value.

Paired vs. Independent Samples The following data represent students’ calculus test scores before and after taking an algebra refresher course. Student 1 2 3 4 5 6 7 8 Before 85 63 94 78 75 82 45 58 After 92 68 98 83 80 88 53 62

Paired vs. Independent Samples Perform a test of the hypotheses H0: 2 – 1 = 0 H1: 2 – 1 > 0 treating the samples as independent.

Paired vs. Independent Samples Had we performed a test of the “same” hypotheses H0: D = 0 H1: D > 0 treating the samples as paired, then the p-value would have been 0.000005688. Why so small?

Paired vs. Independent Samples Why is there a difference? 1 2 3 5 4 6 8 7 40 50 60 80 90 100 70 Paired

Paired vs. Independent Samples Why is there a difference? 1 2 3 5 4 6 8 7 40 50 60 80 90 100 70 Independent

Confidence Intervals Confidence intervals for 1 – 2 use the same theory. The point estimate isx1 –x2. The standard deviation ofx1 –x2 is approximately

Confidence Intervals The confidence interval is or ( known, large samples) ( unknown, large samples) ( unknown, normal pops., small samples)

Confidence Intervals The choice depends on Whether  is known. Whether the populations are normal. Whether the sample sizes are large.

Example Find a 95% confidence interval for 1 – 2 in Example 11.4, p. 699. x1 –x2 = 3.2. sp = 5.052. Use t = 2.101. The confidence interval is 3.2  (2.101)(2.259) = 3.2  4.75.

The TI-83 and Means of Independent Samples To find a confidence interval for the difference between means on the TI-83, Press STAT > TESTS. Choose either 2-SampZInt or 2-SampTInt. Choose Data or Stats. Provide the information that is called for. 2-SampTTest will ask whether to use a pooled estimate of . Answer “yes.”