Independent Samples: Comparing Means Lecture 37 Sections 11.1 – 11.2, 11.4 Tue, Nov 6, 2007

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.

Independent Samples Paired data are often “before” and “after” observations. By comparing the mean before treatment to the mean after treatment, we can determine whether the 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. Furthermore, the samples could be of different sizes.

Independent Samples We start with two populations. Population 1 has mean 1 and st. dev. 1. Population 2 has mean 2 and st. dev. 2. We wish to compare 1 and 2. We do so by taking samples and comparing sample meansx1 andx2.

Independent Samples 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 and comparing it to 0.

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 population of the small sample(s) is normal. in order to use the t distribution.

Further Assumption We will also assume that the two populations have the same standard deviation. Call it . That is,  = 1 = 2. 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 the sample sizes are large enough (or the populations are normal), then according to the Central Limit Theorem, x1 has a normal distribution with mean 1 and standard deviation 1/n1. x2 has a normal distribution with mean 2 and standard deviation 2/n2.

The Distribution ofx1 –x2 It follows from theory thatx1 –x2 is Normal, with Mean Variance

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 Work exercise 11.32 on page 716 under the assumption that  = 6 for both populations.

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 We should use t instead of Z if We use sp instead of , and The sample sizes are small. 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. Treatment 1 is not 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.

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.”