sLecture6bPBS

# sLecture6bPBS - Lecture#6 Learning Objectives 1 Be able to...

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Lecture #6 Learning Objectives 1. Be able to construct confidence interval estimates and to conduct statistical tests about the difference between two populations with respect to the mean when samples are drawn independently from the two populations. ¾ Population variances known ¾ Population variances unknown i Assumed equal ii Assumed unequal 2. Understand the concept and use of a pooled variance estimate. “If you don’t know where you’re going, any road will take you there.” – C.S. Lewis (1898 - 1963)

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BM330 - Lecture 6 Inference About the Difference in Means – Independent Samples In these cases, we wish to determine how two populations differ with respect to the mean of some characteristic, and we collect random samples independently from the two populations. (We could conceivably collect two independent samples from the same population.) Let μ 1 = the mean of population 1 and μ 2 = the mean of population 2. Let 2 1 σ = the variance of population 1 and 2 2 = the variance of population 2. Parameter of interest: μ 1 μ 2 Point Estimator (Sample Statistic): 2 1 X X Sampling Distribution of 2 1 X X : We can use the rules for means and variances reviewed in Lecture 0 to show these are true. Mean of 2 1 X X : 2 1 X X μ = μ 1 μ 2 Variance of 2 1 X X : 2 2 2 1 2 1 2 2 1 n n X X σσ + = Standard Error of 2 1 X X : 2 2 2 1 2 1 2 1 n n X X + = Recall the fact that any linear combination of Normal random variables is also Normal. Therefore, if both of the populations being sampled have a Normal distribution, or both of the samples sizes are sufficiently large, 2 1 X X ~ (at least approximately)Normal. 1
Population Variances Known In this case we assume that, even though we don’t know the values of μ 1 and μ 2 , we do know the values of 2 1 σ and 2 2 . As a result, () () 2 2 2 1 2 1 2 1 2 1 n n X X σσ μμ + = Z ~ N(0, 1) and 2 2 2 1 2 1 * ) ( n n Z σσ + = . Example – Par Golf Company : Par, Inc. is a manufacturer of golf equipment and has recently developed a new golf ball that has been designed for “extra distance.” In a test of driving distance, a sample of par golf balls is compared with a sample of golf balls manufactured by Par's competitor. A mechanical driving device is used to create a constant driving force and the distance for each sample ball hit is recorded. Data: SAMPLE # 1 PAR Golf Balls SAMPLE # 2 Competitor Golf Balls Sample Size n 1 = 120 balls n 2 = 80 balls Sample Mean Distance 1 x = 235 yards 2 x = 218 yards Population Standard Deviation σ 1 = 15 yards σ 2 = 20 yards Calculate the 95% confidence interval estimate of the difference in mean driving distance for Par’s golf ball and the competitor’s golf ball. What do you conclude?

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sLecture6bPBS - Lecture#6 Learning Objectives 1 Be able to...

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