On the basis of data provided by a salary survey, the variance in annual salaries for seniors in public accounting firms is approximately 2.

Question

On the basis of data provided by a salary survey, the variance in annual salaries for seniors in public accounting firms is approximately 2.3 and the variance in annual salaries for managers in public accounting firms is approximately 11.3. The salary data were provided in thousands of dollars. Assuming that the salary data were based on samples of 25 seniors and 26 managers, test to determine whether there is a significant difference between the variances of salaries for seniors and managers. At a 0.05 level of significance, what is your conclusion? State the null and alternative hypotheses.

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Mộc Miên 4 years 2021-08-02T09:07:08+00:00 1 Answers 14 views 0

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  1. trucchi
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    2021-08-02T09:08:08+00:00
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    Answer:

    The null hypothesis is  H_o : \sigma^2_1 = \sigma^2 _2

    The alternative  hypothesis is  H_a : \sigma^2_1 \ne \sigma^2 _2

    The conclusion is  

     There is no  sufficient evidence to conclude that there is a difference between the variances of salaries for seniors and managers

    Step-by-step explanation:

    From the question we are told that

       The variance for seniors is  s ^2_1 = 2.3

       The variance for managers  is s ^2_2 = 11.3

       The  first sample size is  n_1 = 25

        The second sample size is  n_2 = 26

        The significance level is  \alpha = 0.05

         

    The null hypothesis is  H_o : \sigma^2_1 = \sigma^2 _2

    The alternative  hypothesis is  H_a : \sigma^2_1 \ne \sigma^2 _2

    Generally the test statistics is mathematically represented as

          F = \frac{s_1^2}{s_2^2}

    =>    F = \frac{2.3}{11.3}

    =>    F = \frac{2.3}{11.3}

    =>    F = 0.2035

    Generally the p-value is obtain for the F-distribution table (Reference – Free statistic calculator ) at a degrees of freedom

              df_1 = 25 - 1

              df_1 = 24

    and    df_2 = 26 - 1

               df_2 = 25

    The p- value is  

            p-value = f_{0.2035,24,25} = 0.99989

    So from the calculation we see that

           p-value > \alpha

    So we fail to reject the null hypothesis

         

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