Quandaries and Queries
 

 

Ed

Introduction to statistics. Self educator.

This problem is from:

Lipschutz, S. and Schiller, J. Introduction to Probability and Statistics. SchaumÔs 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 PÔ 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 (1-p)/n ) = sqrt ( (0.25 x 0.75)/100 ) = 0.0433. Then

Z = ( PÔ ÷ 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
 
 

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