Question

For each level of precision, find the required sample size to estimate the mean starting salary for a new CPA with 95 percent confidence, assuming a population standard deviation of $7,500 (same as last year).
Required: (Round your answers UP to the nearest integer.)

(a) E = $2,000

Sample size ??????
(b) E = $1,000

Sample size ????
(c) E = $500

Answer 1

Answer:

(a) Margin of error ( E) = $2,000 , n = 54

(b)   Margin of error ( E) = $1,000 , n = 216

(c)   Margin of error ( E) = $500 , n= 864

Step-by-step explanation:

Given -

Standard deviation $\sigma$ = $7,500

$\alpha$ = 1 - confidence interval = 1 - .95 = .05

$Z_{\frac{\alpha}{2}}$ =  $Z_{\frac{.05}{2}}$ = 1.96

let sample size is n

(a) Margin of error ( E) = $2,000

Margin of error ( E)  = $Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

                           E   = $Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}$

Squaring both side

$E^{2} = 1.96^{2}\times\frac{7500^{2}}{n}$

$n =\frac{1.96^{2}}{2000^{2}} \times 7500^{2}$

n =  54.0225

n = 54 ( approximately)

(b)   Margin of error ( E) = $1,000

          E     = $Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

         1000   =  $Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}$

Squaring both side

$1000^{2} = 1.96^{2}\times\frac{7500^{2}}{n}$

$n =\frac{1.96^{2}}{1000^{2}} \times 7500^{2}$

n = 216

(c)   Margin of error ( E) = $500

   E = $Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

  500 = $Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}$

Squaring both side

$500^{2} = 1.96^{2}\times\frac{7500^{2}}{n}$

$n =\frac{1.96^{2}}{500^{2}} \times 7500^{2}$

n = 864