Sec. 5.2 POINT AND INTERVAL ESTIMATION OF p 125 



for which r = 5, 6, 7, 8, 9, or 10. Consequently, if you have drawn 

 a sample with n = 10 and p is unknown, it is quite unlikely that p 

 was as large as 3/4 if it was found in the sample that r — 0, 1, 2, 3, 

 or 4. As a matter of fact, you could just form the habit of assuming 

 that p never is as large as 3/4 whenever r is 0, 1, 2, 3, or 4 and you 

 would be wrong less than 5 per cent of the time because samples of 

 that sort occur less than 5 per cent of the time when n = 10 and p is 

 as large as 3/4. 



To answer the question posed earlier, we consider the following 

 reasoning. If n = 10, the probability series for p set successively 

 equal to 2/3, 3/4, .69, and .70 (for reasons which will appear soon) 

 are obtained as in Chapter 4 and lead to the following conclusions: 



(1/3 + 2/3) 10 = .0000 + .0003 + .0031 + .0163{ + .0569 + .1368 

 (sum = .0197, is <.0250) 



r: 1 2 3 4 5 



+ .2276 + .2601 4- .1951 + .0867}+ .0173 



is <.0250 



r: 6 7 8 9 10 



(Note that the "central 95 per cent" does not include the observed 

 number of occurrences, r = 10.) 



(1/4 + 3/4) 10 = .0000 + .0000 + .0004 + .0031 + .0162{+ .0584 

 (sum = .0197, is <.0250) 



r: 1 2 3 4 5 



+ .1460 4- .2503 + .2816 4- .1877 4- .0563} none ex- 

 cluded 



r: 6 7 8 9 10 



(The central 95 per cent does include the sample result, r = 10, but 

 still might do so with a smaller p; hence p = 3/4 may be too large 

 to be the lower end of the 95 per cent confidence interval. Therefore, 

 p = .70 will be tried.) 



(.3 + .7) 10 = .0000 + .0001 + .0014 + .0090|+ .0368 + .1029 

 (sum = .0105, is <.0250) 



r: 1 2 3 4 5 



+ .2001 + .2668 + .2335 + .1211 4- .0282} none excluded 

 r: 6 7 8 9 10 



