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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
n = number of samples = 40
http://people.hofstra.edu/faculty/St.../SYMB/LSIG.GIF = standard deviation = $9,500
http://people.hofstra.edu/faculty/St...opic1/xbar.gif = mean = $44,000
90% confidence interval = [http://people.hofstra.edu/faculty/St...opic1/xbar.gif - 1.645http://people.hofstra.edu/faculty/St.../SYMB/LSIG.GIF/http://people.hofstra.edu/faculty/St...1/SYMB/SQR.GIFn, http://people.hofstra.edu/faculty/St...opic1/xbar.gif + 1.645http://people.hofstra.edu/faculty/St.../SYMB/LSIG.GIF/http://people.hofstra.edu/faculty/St...1/SYMB/SQR.GIFn]
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?
Are you using t-values or z-values??
I forgot the square root.
n = number of samples = 40
http://people.hofstra.edu/faculty/St.../SYMB/LSIG.GIF = standard deviation = $9,500
http://people.hofstra.edu/faculty/St...opic1/xbar.gif = mean = $44,000
90% confidence interval = [http://people.hofstra.edu/faculty/St...opic1/xbar.gif - 1.645http://people.hofstra.edu/faculty/St.../SYMB/LSIG.GIF/http://people.hofstra.edu/faculty/St...1/SYMB/SQR.GIFn, http://people.hofstra.edu/faculty/St...opic1/xbar.gif + 1.645http://people.hofstra.edu/faculty/St.../SYMB/LSIG.GIF/http://people.hofstra.edu/faculty/St...1/SYMB/SQR.GIFn]
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.
You saved the day :)
