Question

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

The minimum sample size is

(Round up to the nearest integer.)

Answer 1

The minimum sample size required is
`(376.59\ \sigma^(2))`.

Explanation:

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

`CI=\bar x\pm z_(\alpha /2)\ (\sigma)/(√(n))`

The margin of error for this interval is:

`E= z_(\alpha /2)\ (\sigma)/(√(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)\ (\sigma)/(√(n))`

`n=[(z_(\alpha/2)* \sigma)/(E)]^(2)`

`=[(1.96* \sigma)/(0.101)]^(2)\\\\=376.59* \sigma^(2)`

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