Answer:
The 99% confidence interval for the 1999 mean incomes of Middletown householders is ($33,013.5, $33,892.5).
Step-by-step explanation:
Of the 40,000 persons, 2621 householders reported their income.
The sample size of the study is, n = 2621.
As the sample size is large, i.e. n > 30 use z-distribution to compute the confidence interval.
The confidence interval for mean is:
[tex]CI=\bar x\pm z_{\alpha /2}\frac{\sigma}{\sqrt{n}}[/tex]
The mean of this sample is, [tex]\bar x=\$33,453[/tex]
The standard deviation is,[tex]s\approx\sigma=\$8,721[/tex]
The critical value of z for 99% confidence level is:
[tex]z_{\alpha /2}=z_{0.01/2}=z_{0.005}=2.58[/tex]
The 99% confidence interval for the 1999 mean incomes of Middletown householders is:
[tex]CI=\bar x\pm z_{\alpha /2}\frac{\sigma}{\sqrt{n}}\\=33453\pm2.58\times\frac{8721}{\sqrt{2621}}\\=33453\pm439.50\\=(33013.5, 33892.5)[/tex]
Thus, the 99% confidence interval for the 1999 mean incomes of Middletown householders is ($33,013.5, $33,892.5).
