5.8. In Problem 5.6, p-bar0.1064 was found by use of (5.8), that is, 1873=17,608. Do you think...

5.8. In Problem 5.6, p-bar¼0.1064 was found by use of

(5.8), that is, 1873=17,608. Do you think you would obtain the same p by adding the 24 p values and dividing by 24?

This would be an unweighted average of p’s. If we substitute np for d in (6.8) we have (

P np)=

P n. Thus p is a weighted average of p values with weights n.