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 March 20th, 2007, 05:52 AM #1 Senior Member   Joined: Dec 2006 From: New Jersey Posts: 378 Thanks: 1 Confidence Intervals f1) A study was conducted to estimate the mean amount spent on birthday gifts for a typical family having two children. A sample of 150 was taken, and the mean amount spent was $225. Assuming a standard deviation equal to$50, find the 95% confidence interval for the population mean, the mean for all such families. 2) What sample size would be needed to estimate the population mean to within one-half standard deviation with 95% confidence? .
 March 29th, 2007, 08:59 PM #2 Member   Joined: Mar 2007 Posts: 57 Thanks: 0 1) Given n = 150 sample mean = y bar = 225 assuming the stdev of 50 given is the population stdev and not the sample stdev, i.e. sigma = 50 100(1-a)=95% so a = 5% (alpha) z_(a/2) = z_0.025 = 1.96 Then the 95% CI for mu is: ( y bar - z_a/2 * sigma/ n^(1/2) , y bar + z_a/2 * sigma/ n^(1/2) ) = ( 225 - 1.96 * 50/150^(1/2) , 225 + 1.96 * 50/150^(1/2) ) = ( 217.00, 233.00) 2) From before the 95% CI for mu is: ( y bar - z_a/2 * sigma/ n^(1/2) , y bar + z_a/2 * sigma/ n^(1/2) ) one half stdev = 1/2 * 50 = 25 so we want 2 * (z_a/2 * sigma/ n^(1/2)) <25>= (2*1.96*50/25)^2 = 61.4656 So the sample size needed is at least 62. note: I may be wrong.

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# 95% confidence interval for the population mean amount spent

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