# ch6 - AMS 315/576 Lecture Notes Chapter 6. Inference about...

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¤ § ¥ ƒ AMS 315/576 Lecture Notes Chapter 6. Inference about μ 1 - μ 2 1. Large Populations ( n 1 30 , n 2 30 ) Two Independent Samples / £ ¡ ¢ EXAMPLE 6.1 A survey of credit card holders revealed that Americans carried an average credit card balance of \$3900 in 1995 and \$3300 in 1994 ( U.S. News & World Report , January 1, 1996). Suppose that these averages are based on random samples of 400 credit card holders in 1995 and 450 credit card holders in 1994 and that the population standard deviations of the balances were \$880 in 1995 and \$810 in 1994. (a) What is the point estimate of μ 1 - μ 2 ? (b) Construct a 95% conﬁdence interval for the diﬀerence between the mean credit card balances for all credit card holders in 1995 and 1994. SOLUTION Refer to all credit card holders in 1995 as population 1 and all credit card holders in 1994 as population 2. The respective samples, then, are samples 1 and 2. Let μ 1 and μ 2 be the mean credit card balances for populations 1 and 2, and let ¯ x 1 and ¯ x 2 be the means of the respective samples. From the given information, For 1995: n 1 = 400 , ¯ x 1 = \$3900 , σ 1 = \$880 For 1994: n 2 = 450 , ¯ x 2 = \$3300 , σ 2 = \$810 (a) The point estimate of μ 1 - μ 2 is given by the value of ¯ x 1 - ¯ x 2 . Thus, Point estimate of μ 1 - μ 2 = \$3900 - \$3300 = \$600 (b) The conﬁdence level is 1 - α = . 95. First, we calculate the standard deviation of ¯ x 1 - ¯ x 2 as follows. σ ¯ x 1 - ¯ x 2 = s σ 2 1 n 1 + σ 2 2 n 2 = r (880) 2 400 + (810) 2 450 = \$58 . 25804665 Next, we ﬁnd the z value for the 95% conﬁdence level. From the normal distribution table, this value of z is 1.96. 1

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Finally, substituting all the values in the conﬁdence interval formula, we obtain the 95% conﬁdence interval for μ 1 - μ 2 as x 1 - ¯ x 2 ) ± ¯ x 1 - ¯ x 2 = (3900 - 3300) ± 1 . 96(58 . 25804665) = 600 ± 114 . 19 = \$ 485 . 81 to \$ 714 . 19 Thus, with 95% conﬁdence we can state that the diﬀerence in the mean credit card balances for all credit card holders in 1995 and 1994 was be- tween \$485.81 and \$714.19. / £ ¡ ¢ EXAMPLE 6.2 In 1996, Money magazine conducted an experiment in which one-dollar coins and quarters were left on busy sidewalks in six major U.S. cities ( Money , October 1996). An observer then recorded the length of time each coin remained on the sidewalk before being picked up by a pedestrian. According to the results of the experiment, the mean length of time one-dollar coins remained on the sidewalks before being picked up by pedestrians was 6.50 minutes and the mean length of time quarters stayed on the sidewalks was 5.75 minutes. Assume that these means are based on samples of 50 one-dollar coins and 45 quarters and that the two sample standard deviations are 1.75 minutes and 1.20 minutes, respectively. Find a 99% conﬁdence interval for the diﬀerence between the corresponding population means. / £ ¡ ¢ EXAMPLE 6.3 Refer to Example 6.2 about the mean length of times one-dollar coins and quarters remained on the sidewalks before being picked up by pedestrians. Test at the 2.5% signiﬁcance level if the mean length of time all one-dollar coins will remain on the sidewalks before being picked up is higher than the mean length of time all quarters will stay on the sidewalks.
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## This note was uploaded on 01/31/2011 for the course AMS 572 taught by Professor Weizhu during the Fall '10 term at SUNY Stony Brook.

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ch6 - AMS 315/576 Lecture Notes Chapter 6. Inference about...

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