Question

uppose a large telephone manufacturer has a problem with excessive customer complaints and consequent returns of the phones for repair or replacement. The manufacturer wants to estimate the magnitude of the problem in order to design a quality control program. How many telephones should be sampled and checked in order to estimate the proportion defective to within 4 percentage points with 83% confidence

Answer 1

Answer:

The answer is "294.2075".

Step-by-step explanation:

Given:

$\hat{p} = 0.5 \\\\1 - \hat{p} = 1 - 0.5 = 0.5\\\\E = 4\% = 0.04$

At $83\%$ confidence level the z is ,

$\alpha = 1 - 83\% = 1 - 0.83 = 0.17\\\\\frac{\alpha}{2} = \frac{0.17}{2} = 0.085\\\\Z_{\frac{\alpha}{2}} = Z_{0.085} = 1.3722\\\\n = (\frac{Z_{\frac{\alpha}{2}}}{E})^2 \times \hat{p} \times (1 - \hat{p})$

   $= (\frac{1.3722}{0.04})^2 \times 0.5 \times 0.5\\\\= (34.305)^2 \times 0.5 \times 0.5\\\\= 1,176.83 \times 0.5 \times 0.5\\\\= 1,176.83 \times 0.25\\\\=294.2075$