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, <em>n</em> = 2621.
As the sample size is large, i.e. <em>n</em> > 30 use <em>z</em>-distribution to compute the confidence interval.
The confidence interval for mean is:
![CI=\bar x\pm z_{\alpha /2}\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=CI%3D%5Cbar%20x%5Cpm%20z_%7B%5Calpha%20%2F2%7D%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
The mean of this sample is, ![\bar x=\$33,453](https://tex.z-dn.net/?f=%5Cbar%20x%3D%5C%2433%2C453)
The standard deviation is,![s\approx\sigma=\$8,721](https://tex.z-dn.net/?f=s%5Capprox%5Csigma%3D%5C%248%2C721)
The critical value of <em>z</em> for 99% confidence level is:
![z_{\alpha /2}=z_{0.01/2}=z_{0.005}=2.58](https://tex.z-dn.net/?f=z_%7B%5Calpha%20%2F2%7D%3Dz_%7B0.01%2F2%7D%3Dz_%7B0.005%7D%3D2.58)
The 99% confidence interval for the 1999 mean incomes of Middletown householders is:
![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)](https://tex.z-dn.net/?f=CI%3D%5Cbar%20x%5Cpm%20z_%7B%5Calpha%20%2F2%7D%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%5C%5C%3D33453%5Cpm2.58%5Ctimes%5Cfrac%7B8721%7D%7B%5Csqrt%7B2621%7D%7D%5C%5C%3D33453%5Cpm439.50%5C%5C%3D%2833013.5%2C%2033892.5%29)
Thus, the 99% confidence interval for the 1999 mean incomes of Middletown householders is ($33,013.5, $33,892.5).