### Ch. 7, #14

(a) In other words, we want the 90% confidence interval to be plus or minus 1.0 from the mean. The standard error of the mean is 3.8/sqrt(n). The 5% one-tail critical value is 1.645 from either the standard normal table or the bottom row of the t table. (Please look up both so you understand this point.)

So, we want 1.645 x 3.8/sqrt(n) = 1.0. Or, sqrt(n) = 1.645 x 3.8 = 6.2510. Squaring shows that the required n is 39.075, whick mean n = 40 to strict inequalities

(b) To switch to a 95% confidence interval, we would change 1.645 to 1.96. The required n is found by sqrt(n) = 1.96 x 3.8 = 7.446. This gives us n = 55.47 or 56 observations.

(c) If we relax the requirement to a confidence interval of plus or minus 1.5 we have 1.96 x 3.8/sqrt(n) = 1.5 or sqrt(n) = 1.96 x 3.8 / 1.5. So, sqrt(n) = 4.9653. The required n is 24.65 or 25 observations.

Note that here I did the calculations. I am also not real impressed by the answer to (c) in the back of the book.

Posted by bparke at November 13, 2003 11:33 AM