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 \(z_{\alpha/2}\) corresponding to the area of \(1-\alpha/2 = 0.95\) in the standard normal distribution is \(z_{0.05}\approx1.645\) (from standard normal table).

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

The formula for the margin of error when the population standard deviation \(\sigma\) is known is \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\).
Given \(n = 50\), \(\sigma=630\), \(z_{\alpha/2}=1.645\)
\(E=1.645\times\frac{630}{\sqrt{50}}\)
First, \(\sqrt{50}\approx7.071\)
\(E = 1.645\times\frac{630}{7.071}\)
\(\frac{630}{7.071}\approx89.1\)
\(E=1.645\times89.1\approx146.6\)

Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is given by \(\bar{x}-E<\mu <\bar{x} + E\)
Given \(\bar{x}=3987\)
\(\bar{x}-E=3987- 146.6=3840.4\approx3840\)
\(\bar{x}+E=3987 + 146.6=4133.6\approx4134\)

Answer:

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