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6 August, 07:28

Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $43,000 and $61,600. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired.

Required:

a. What is the planning value for the population standard deviation (to the nearest whole number) ?

b. How large a sample should be taken if the desired margin of error is as shown below (to the nearest whole number) ?

1. $500?

2. $200?

3. $100?

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Answers (1)
  1. 6 August, 11:24
    0
    a. 4650

    b.

    1. 332

    2. 2076

    3. 8306

    Step-by-step explanation:

    a. The planning value for population standard deviation is,

    s = (maximum - minimum) / 4

    s = (61600 - 43000) / 4

    s = 4650

    that is, it would be 4650, the planning value for population standard deviation

    b. we have to:

    n = (z * s / E) ^ 2

    z for confidence level 95% is 1.96, E = 500; 200; 100

    replacing:

    1.

    n = (1.96 * 4650/500) ^ 2

    n = 332.2 = 332

    2.

    n = (1.96 * 4650/200) ^ 2

    n = 2076.6 = 2076

    3.

    n = (1.96 * 4650/100) ^ 2

    n = 8306.4 = 8306
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