LL6 - 2 12 2 that X 1 X 2 follows N 1 2 n1 n2 Therefore X 1 X 2 is an unbiased point estimator of 1 2 Part 6 Comparison of Two Populations 1

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
BUS1200-6 1 Part 6 Comparison of Two Populations 1. Inferences about the Difference between Two Normal Population Means: Independent Samples 2. Inferences about the Difference between Two Normal Population Means: Paired Samples Section 6.1 Inferences about the Difference between Two Normal Population Means: Independent Samples Assumption The two populations follow N( µ 1 , σ 1 2 ) and N( 2 , 2 2 ) respectively. Two random samples, one from each of the two populations, are drawn independently . Let n 1 and n 2 be the two sample sizes, 1 X and 2 X be the two sample means, and S 1 2 and S 2 2 be the two sample variances. It can be proved BUS1200-6 2 that 2 1 X X follows + 2 2 2 1 2 1 2 1 , N n n . Therefore 2 1 X X is an unbiased point estimator of 1 2 . Case 1: 1 2 and 2 2 are known Interval estimation The 100(1 α )% confidence interval for 1 2 is + + + 2 2 2 1 2 1 2 / 2 1 2 2 2 1 2 1 2 / 2 1 , n n z x x n n z x x . Example 1 Construction workers earned an average of $551 per week and manufacturing workers earned an average of $487 per week. The above mean earnings are actually obtained from samples of 500 and 700 workers taken from the two groups of workers respectively. Further assume that the standard deviations of weekly earnings of the two groups of workers are $66 and $60 respectively. Construct a 95% confidence BUS1200-6 3 interval for the difference between the mean weekly earnings of the two groups of workers. [Solution] Refer to weekly earnings of all construction workers as population 1 and those of all manufacturing workers as population 2. The respective samples are then samples 1 and 2. From the given information, n 1 = 500, 1 x = 551, 1 = 66, n 2 = 700, 2 x = 487, 2 = 60, = 0.05. 2 1 x x = 551 487 = 64, 700 60 500 66 960 . 1 2 2 2 2 2 1 2 1 2 / + = + n n z = 7.30, 64 7.30 = 56.70, 64 + 7.30 = 71.30. Thus, the 95% confidence interval for the difference in the mean weekly earnings of all construction workers and all manufacturing workers is ($56.70, $71.30). Hypothesis test The test statistic is () 2 2 2 1 2 1 2 1 n n X X + . Let be the level of significance. BUS1200-6 4 H 0 H 1 Reject H 0 if and only if 1 = 2 or 1 2 1 < 2 2 2 2 1 2 1 2 1 n n x x + −≤ z 1 = 2 or 1 2 1 > 2 2 2 2 1 2 1 2 1 n n x x + −≥ z 1 = 2 1 2 2 2 2 1 2 1 2 1 n n x x + z / 2 Example 2 Reconsider the previous example. Test at the 1% significance level if the mean weekly earnings of the two groups of workers are significantly different. [Solution] H 0 : 1 = 2 , H 1 : 1 2 , = 0.01. The value of the test statistic is 700 60 500 66 487 551 2 2 2 2 2 1 2 1 2 1 + = + n n x x = 17.19. Since |17.19| 2.576 = z 0.005 , we reject H 0 . Thus we conclude that there is a significant
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
BUS1200-6 5 difference between the mean weekly earnings of the two groups of workers at the 1% significance level. Case 2: σ 1 2 and 2 2 are unknown but equal Interval estimation Let S p = 2 ) 1 ( ) 1 ( 2 1 2 2 2 2 1 1 + + n n S n S n be the pooled sample standard deviation . S p 2 is called the pooled sample variance , and is an unbiased estimator of the common value of 1 2 and 2 2 .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/03/2011 for the course ECON 101 taught by Professor Wood during the Spring '07 term at University of California, Berkeley.

Page1 / 5

LL6 - 2 12 2 that X 1 X 2 follows N 1 2 n1 n2 Therefore X 1 X 2 is an unbiased point estimator of 1 2 Part 6 Comparison of Two Populations 1

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online