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    Hypothesis testing

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    Every year the Wall Street Journal has a feature on compensation of CEOs of different companies. The data include type of company, amount of salary, amount of bonus, % change from the previous year, and several other compensation forms. A sample of the data is shown in the table. (SEE ATTACHED TABLE!)

    a). Look at the 1996 salary data for cyclical and noncyclical companies. Create a plot of salaries for each group. Based on the graphs do you think the data are normally distributed? Why or why not?

    b). Perform the appropriate hypothesis test to determine whether the variances in salary for the two types of companies are equal. Based on this test, can you assume equal variances?

    c). Based on the results of your answers to parts (a) and (b) select the appropriate test procedure to test whether or not the mean salary for cyclical and noncyclical companies is the same.

    d). Perform the test at the 0.05 level of significance. What is your conclusion?

    e). Repeat the procedure you used to answer parts (a) to (d) to determine whether the mean bonus for the two types of companies is different.

    f). Look at the entire set of data. One variable that is reported is the change in compensation level from the previous year (%). Calculate the proportion of companies whose CEOs received increases in compensation. Do the same thing for the proportion who received decreases in compensation.

    g). At the 0.05 level of significance, is the proportion of those receiving positive increases greater than the proportion of those receiving decreases?

    h). Do the necessary analysis to compare the mean salaries of the technology companies to those of the industrial companies.

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    Cyclical companies:
    Alumax : Architectural products
    Armco : Real estate
    Because, their business reacts to economic change. People spend money more conservatively in recessions.

    Noncyclical companies:
    Air products

    Let salary of cyclical companies = x and of noncyclical = y
    <x> = (800+559.2)/2 = 679.6 thousand dollars
    n1 = 2
    sx = RMS(x) = sqrt(((800 - 679.6)^2 + (559.2 - 679.6)^2)/(2-1))
    => sx = 170.27
    Check for normality:
    It's very difficult to confirm the normality with small sample.
    <x> - sx = 679.6-170.27 = 509.33
    <x> + sx = 679.6+170.27 = 849.87
    no. of observations out side (<x> - sx,<x> + sx) = 0 => probability = 0 (theoretically 1/3)
    hence all the data are within (<x> - sx,<x> + sx), i.e., probability = 1 (theoretically 2/3).
    Here, it's bit difficult to decide the normality.
    see attached file, it is symmetric about mean (<x>) so we can assume it to be normally distributed.
    <y> = (738.1 + 750 + 811.7)/3 = 766.6 thousand dollars

    n2 =3
    sy = sqrt(((738.1-766.6)^2 + (750-766.6)^2 + (811.7-766.6)^2)/(3-1))
    => sy = 39.508 = 39.51
    <y> - sy = 766.6 - 39.51 = 727.09
    <y> + sy = 766.6 ...

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