Question

Find the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income.

Answer 1

The smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income is 48.

Explanation:

The complete question is:

The mean salary of people living in a certain city is $37,500 with a standard deviation of $2,103. A sample of n people will be selected at random from those living in the city. Find the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income. Round your answer up to the next largest whole number.

Solution:

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

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

The margin of error for this interval is:

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

The critical value of z for 90% confidence level is:

z = 1.645

Compute the required sample size as follows:

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

`n=[(z_(\alpha/2)\cdot\sigma)/(MOE)]^(2)\\\\=[(1.645* 2103)/(500)]^(2)\\\\=47.8707620769\\\\\approx 48`

Thus, the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income is 48.