Question

Problem 2: It has been reported that 40% of U.S. workers employed as purchasing managers are females. In a simple random sample of U.S. purchasing managers, 70 out of the 200 are females. Given this information: Source: Bureau of the Census, Statistical Abstrad of the United States 2009, p. 384. a. What is the population proportion, as b. What is the sample proportion, poster c. What is the standard error of the sample proportion? d. In the sampling distribution of the proportion, what is the probability that a sample of this size would result in a sample proportion at least as large as that found in part (b)?!

Answer 1

a)

Population proportion $\pi$ = 40% or 0.4

It is the percentage of US workers employed as purchasing managers who are females.

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b)

Sample proportion p = ( 70 $\div$ 200) = 35% or 0.35

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c)

Standard error of the sample proportion is calculated using the following equation

0.4 x (1– 0.4) Op = = 0.0346 %3D 200

Standard error of the sample proportion = 0.0346

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d)

p - Op d,

0.35 – 0.40 0.0346

Z = -1.45

p( z $\leq$ -1.45) = 0.07353

Probability that sample of this size would result in sample proportion at least as large as 0.35 = 0.07353