Question

Estimate the minimum sample size needed to achieve the margin of error E= 0.101 for a 95% confidence interval.

Answer 1

Answer:

The minimum sample size required is $(376.59\ \sigma^{2})$.

Step-by-step explanation:

The (1 - α) % confidence interval for population mean is:

 $CI=\bar x\pm z_{\alpha /2}\ \frac{\sigma}{\sqrt{n}}$

The margin of error for this interval is:

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

The information provided is:

 E = 0.101

Confidence level = 95%

α = 5%

Compute the critical value of z for α = 5% as follows:

 $z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use a z-table.

Compute the sample size required as follows:

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

       $n=[\frac{z_{\alpha/2}\times \sigma}{E}]^{2}$

          $=[\frac{1.96\times \sigma}{0.101}]^{2}\\\\=376.59\times \sigma^{2}$

Thus, the minimum sample size required is $(376.59\ \sigma^{2})$.