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.

Respuesta :

Answer:

The null hypothesis is  [tex]H_o : \sigma^2_1 = \sigma^2 _2[/tex]

The alternative  hypothesis is  [tex]H_a : \sigma^2_1 \ne \sigma^2 _2[/tex]

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  [tex]s ^2_1 = 2.3[/tex]

   The variance for managers  is [tex]s ^2_2 = 11.3[/tex]

   The  first sample size is  [tex]n_1 = 25[/tex]

    The second sample size is  [tex]n_2 = 26[/tex]

    The significance level is  [tex]\alpha = 0.05[/tex]

     

The null hypothesis is  [tex]H_o : \sigma^2_1 = \sigma^2 _2[/tex]

The alternative  hypothesis is  [tex]H_a : \sigma^2_1 \ne \sigma^2 _2[/tex]

Generally the test statistics is mathematically represented as

      [tex]F = \frac{s_1^2}{s_2^2}[/tex]

=>    [tex]F = \frac{2.3}{11.3}[/tex]

=>    [tex]F = \frac{2.3}{11.3}[/tex]

=>    [tex]F = 0.2035[/tex]

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

          [tex]df_1 = 25 - 1[/tex]

          [tex]df_1 = 24[/tex]

and    [tex]df_2 = 26 - 1[/tex]

           [tex]df_2 = 25[/tex]

The p- value is  

        [tex]p-value = f_{0.2035,24,25} = 0.99989[/tex]

So from the calculation we see that

       [tex]p-value > \alpha[/tex]

So we fail to reject the null hypothesis