Quandaries
and Queries 

Ed Introduction to statistics. Self educator. This problem is from: Lipschutz, S. and Schiller, J. Introduction to Probability and Statistics. Schaumâs Outlines. 1998. Page 217 Example 7.10 Suppose 25 percent of all U.S. workers belong to a labor union. What is the probability that in a random sample of 100 U.S. workers, at least 20 percent will belong to a labor union? Solution: The sample size, n = 100, is greater than 30, and the total number of U.S. workers is much larger than 100. Therefore, the sample proportion Pâ of workers that belong to a labor union can be modeled as a normal random variable with mean p = 0.25, and standard deviation My notation for typing: Let sqrt = square root, Let SD = standard deviation, Let SE standard error. Back to the problem. SD = sqrt( p (1p)/n ) = sqrt ( (0.25 x 0.75)/100 ) = 0.0433. Then Z = ( Pâ ö 0.25 )/ 0.0433 Is a standard normal variable. Using the standard normal table, P( P? >= 0.2 ) = P ( (P?  0.25)/0.0433 >= (0.2  0.25)/0.0433 = P( Z >= 1.15 ) = P( Z <= 1.15) = 0.8749 End their solution. Here is my problem 1: When I use the table to obtain P( Z <= 1.15 ) I get 0.7498 as the area between Z and Z. That leaves 1  0.7498 as the area to the left of Z, NOT 0.8749. So where does the 0.8749 come from? Here is my problem 2: They use P( P? >= 0.2 ) = (0.2  0.25)/0.0433. I thought Z is the (observed  expected)/SD. Shouldn't it be (0.25  0.2)/0.433? 

Hi Ed, Your table gives 0.7498 ad the probability that Z is between 1.15 and 1.15. Half of that, that is, 0.3749, is the probability that Z is between 0 and 1.15. You then add the 0.5000 probability that Z is negative to get 0.8749. For your second question "expected" is in the whole population: 25%, or 0.25. "observed" is a bit of a misnomer here. The 20% is the lower bound of the proportion of unionized worker in your sample that would give an affirmative answer to the question "Are at least 20% of the workers in my sample in a union?" Claude 

