Question

What is the minimum sample size required to estimate a population mean with 90% confidence when the desired margin of error is D

Answer 1

Answer:

The expression for the required minimum sample size is $n=[\frac{1.645\times\sigma }{D}]^{2}$.

Step-by-step explanation:

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

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

The margin of error for this interval is:

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

Confidence level = 90%

α = 10%

Compute the critical value of z for α = 10% as follows:

 z = 1.645

*Use a z-table.

Compute the sample size required as follows:

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

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

$n=[\frac{1.645\times\sigma }{D}]^{2}$

Thus, the expression for the required minimum sample size is $n=[\frac{1.645\times\sigma }{D}]^{2}$.