Respuesta :
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:
[tex]CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}[/tex]
The margin of error for this interval is:
[tex]MOE=z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}[/tex]
The critical value of z for 90% confidence level is:
z = 1.645
Compute the required sample size as follows:
[tex]MOE=z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}[/tex]
[tex]n=[\frac{z_{\alpha/2}\cdot\sigma}{MOE}]^{2}\\\\=[\frac{1.645\times 2103}{500}]^{2}\\\\=47.8707620769\\\\\approx 48[/tex]
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.
