Independent Samples: Comparing Means Lecture 48 Section 11.4 Tue, Apr 20, 2004

Independent Samples If the samples are independent, then there is no sensible way to pair the data. Indeed, the sample sizes may be different. If we wish to compare population means 1 and 2, we must compare sample meansx1 and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 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 . This assumption should be supported by the evidence.

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.

Some Theory Let X1 be a random value from population #1 and X2 a random value from population #2. Then The mean of X1 – X2 is 1 – 2. The variance of X1 – X2 is 12 + 22. The standard deviation of X1 – X2 is (12 + 22).

Some More Theory If X1 and X2 are normal, then so is X1 – X2. Therefore, If X1 is N(1, 1) and X2 is N(2, 2), Then X1 – X2 is N(1 – 2, (12 + 22)).

Consequence of Theory Apply this theory tox1 andx2. We know that x1 is N(1, /n1) and x2 is N(2, /n2). Therefore, x1 –x2 is

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

Hypothesis Testing See Example 11.4, p. 647 – Comparing Two Headache Treatments. State the hypothesis. H0: 1 = 2 H1: 1 > 2 State the level of significance.  = 0.05.

The t Statistic, a.k.a. Our Second Bad Formula Compute the test statistic. The test statistic is

Computations

Hypothesis Testing Calculate the p-value. The number of degrees of freedom is df = df1 + df2 = 18. p-value = P(t(18) > 1.416) = tcdf(1.416, 99, 18) = 0.0869.

Hypothesis Testing State the conclusion. Since p-value > , we conclude that “at the 5% level of significance, we cannot reject the claim that Treatment 1 is as effective as Treatment 2.”

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 sp.

Confidence Intervals The confidence interval is (x1 –x2 )  z  (1/n1 + 1/n2) or (x1 –x2 )  z  sp (x1 –x2 )  t  sp

Confidence Intervals The choice depends on The sample sizes. Whether  is known.

Example Find a 95% confidence interval for 1 – 2 in Example 11.4. 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.

Assignment Page 660: Exercises 17 – 20, 22, 23, 24*, 25*, 28.