Question

round answers to whole numbers. write confidence interval using $ sign as an ordered pair with space after the comma.
a random sample of 50 four - year - olds attending day care centers provided a yearly tuition average of $3987 and the population standard deviation of $630. find the 90% confidence interval of the true mean. if a day care center were starting up and wanted to keep tuition low, what would be a reasonable amount to charge?

Explanation:

Step1: Find the z - score for 90% confidence level

For a 90% confidence level, the significance level \(\alpha=1 - 0.90 = 0.10\), and \(\alpha/2=0.05\). The z - score \(z_{\alpha/2}\) is such that \(P(Z>z_{\alpha/2}) = 0.05\) and \(P(Z < z_{\alpha/2})=0.95\). From the standard normal table, \(z_{\alpha/2}=z_{0.05}\approx1.645\)

Step2: Calculate the margin of error \(E\)

The formula for the margin of error for a population mean when the population standard deviation \(\sigma\) is known is \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\)
Given \(n = 50\), \(\sigma=630\), and \(z_{\alpha/2}=1.645\)
\(E=1.645\times\frac{630}{\sqrt{50}}\)
\(E = 1.645\times\frac{630}{7.071}\)
\(E=1.645\times89.1\)
\(E\approx146.6\)

Step3: Calculate the confidence interval

The formula for the confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu <\bar{x} + E\)
Given \(\bar{x}=3987\)
\(\bar{x}-E=3987- 146.6=3840.4\)
\(\bar{x}+E=3987 + 146.6=4133.6\)

Answer:

\((\$3840,\$4134)\)