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March 20th, 2007, 06:52 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 onehalf standard deviation with 95% confidence? . 
March 29th, 2007, 09:59 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(1a)=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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