1. ## Hard Problem

A sample of 40 families is taken, and a sample mean income of $44,000 with a sample standard deviation of$9,500 is calculated. Determine a 90% confidence interval for the population means income.

a. $41,469, 46,531 b.$41,529, $46,471 c.$40,233, $45,887 d.$45,676, $50,324 2. n = number of samples = 40 = standard deviation =$9,500

= mean = $44,000 90% confidence interval = [ - 1.645/n, + 1.645/n] the number 1.645 from the above equation is looked up in a table or in a calculator I can show you how to do this if you need it. Therefore the interval is 44,000 plus or minus (1.645*9,500/40) somewhow I am not getting an answer in your list . . . can anyone help? 3. Are you using t-values or z-values?? 4. I forgot the square root. n = number of samples = 40 = standard deviation =$9,500
= mean = $44,000 90% confidence interval = [ - 1.645/n, + 1.645/n] the number 1.645 from the above equation is looked up in a table or in a calculator I can show you how to do this if you need it. Therefore the interval is 44,000 plus or minus (1.645*9,500/sqrt(40)) this gives: b.$41,529, \$46,471 as the answer.

5. ## Thanks

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