Question

A statistics practitioner would like to estimate a population mean to within 50 units with 99% confidence given that the population standard deviation is 250. What sample size should be used? b. Re-do part (a) changing the standard deviation to 50. c. Re-do part (a) using a 95% confidence leve

Answer 1

(a) 167

(b) 7

(c) 97

Explanation:

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

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

The margin of error is:

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

Then the formula to estimate the sample size is:

`n=[(z_(\alpha/2)\cdot \sigma )/(MOE)]^(2)`

(a)

For 99% confidence interval the critical value of z is:

z = 2.58.

The standard deviation is, 250.

Compute the sample size as follows:

`n=[(z_(\alpha/2)\cdot \sigma )/(MOE)]^(2)`

`=[(2.58* 250)/(50)]^(2)\\\\=(12.9)^(2)\\\\=166.41\\\\\approx 167`

The sample size that should be used is 167.

(b)

Now the standard deviation is, 50.

Compute the sample size as follows:

`n=[(z_(\alpha/2)\cdot \sigma )/(MOE)]^(2)`

`=[(2.58* 50)/(50)]^(2)\\\\=(2.58)^(2)\\\\=6.6564\\\\\approx 7`

The sample size that should be used is 7.

(c)

Now a 95% confidence level is used.

For 95% confidence interval the critical value of z is:

z = 1.96.

Compute the sample size as follows:

`n=[(z_(\alpha/2)\cdot \sigma )/(MOE)]^(2)`

`=[(1.96* 250)/(50)]^(2)\\\\=(9.8)^(2)\\\\=96.04\\\\\approx 97`

The sample size that should be used is 97.