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6 March, 19:34

A sample of 75 information systems managers had an average hourly income of $40.75 with a standard deviation of $7.00. The 95% confidence interval for the average hourly wage (in $) of all information system managers is

a. 37.54 to 43.96.

b. 38.61 to 42.89.

c. 39.14 to 42.36.

d. 39.40 to 42.10.

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  1. 6 March, 21:32
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    Answer: c. 39.14 to 42.36

    Step-by-step explanation:

    We want to determine a 95% confidence interval for the average hourly wage (in $) of all information system managers

    Number of sample, n = 75

    Mean, u = $40.75

    Standard deviation, s = $7.00

    For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

    We will apply the formula

    Confidence interval

    = mean + / - z * standard deviation/√n

    It becomes

    40.75 + / - 1.96 * 7/√75

    = 40.75 ± 1.96 * 0.82

    = 40.75 + 1.6072

    The lower end of the confidence interval is 40.75 - 1.6072 = 39.14

    The upper end of the confidence interval is 40.75 + 1.6072 = 42.36
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