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A Social service agency plans to conduct a survey to determine the mean income of its clients. The director of the agency prefers that you measure the mean income very accurately, to within plus or minus $500. from a sample taken 2 years ago, you estimate that the standard deviation of income for this population is about $5000. I need to figure out the necessary sample size to reduce the confidence interval to plus/minus$500

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okpalawalter8

Answer:

the necessary sample size to reduce the confidence interval to plus/minus$500 is

                               [tex]n \approx 385[/tex]

         

Step-by-step explanation:

We are given from the question that

          Marginal error E = $500

        The standard deviation is [tex]\sigma[/tex] = $5000

        Let the level of significance be [tex]\alpha = 0.05[/tex]

In order to obtain the sample size n we make use of formula

                       [tex]n = [\frac{Z_{\alpha /2}S}{E}]^2[/tex]

 Now  [tex]{Z_{\alpha /2}[/tex] is the Z score of [tex]\alpha[/tex]/2 and this can be obtained from the z-table and from the z table the  z - score for this case of level of significance  is 1.96

  Substituting value into the formula above

                 [tex]n = (\frac{(1.96* 5000)}{500})^2[/tex]

                [tex]n \approx 385[/tex]

         

             

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