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

Answer:

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.

Step-by-step 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:

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

The margin of error for this interval is:

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

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

z = 1.645

Compute the required sample size as follows:

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

      $n=[\frac{z_{\alpha/2}\cdot\sigma}{MOE}]^{2}\\\\=[\frac{1.645\times 2103}{500}]^{2}\\\\=47.8707620769\\\\\approx 48$

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.