Hello community, Let me first state that I teach myself statistics with some books. I am not shure if there is a typo in the book I am using or I am mistaken with the solution approach or misinterprete the language (my mothertongue is german).
Jerrold H. Zar, Biostatistical Analysis, 2nd edition, Chapter 8, Exercise 8.4, (page 120) A sample of size 18 has a mean of 13.55cm and a variance of 6.4512cm^2. (a) Calculate the 95% confidence interval for the population mean. (b) How large a sample would have to be taken from this population to estimate mu to within 1.00cm, with 95% confidence? (c) 2.00cm with 95% confidence? (d) 2.00cm with 99% confidence? (a) is 13.55cm +- t(0.05,2,17) * sqrt(variance) 13.55cm +- 1.26cm, no problem with this. Here comes my problem: According to the answers appendix of the book (b) n=29, (c) n=10, (d) n=11 I use Excel to iterate n using the formula n= s^2 * t^2 alpha(2),(n-1) * F beta(1),(n-1,nu)/d^2 in Excel: n=s^2*TINV(alpha,n-1)^2 * FINV(beta,n-1,nu)/d^2 For this example: nu=17, alpha= 0.05, d=0.5cm Where is beta in (b), (c) and (d) ? I thought it would be defined by stating (b) "within 1.00cm" which means an accuracy. "1.00" implies a range of accuracy of 0.01 cm. So beta= 0.01. And n would be 270. . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
