Question

8.10 construct a confidence interval for μ assuming that each sample is from a normal population.
a. (\bar{x}=24), (sigma = 3), (n = 10), 90 percent confidence
b. (\bar{x}=125), (sigma = 8), (n = 25), 99 percent confidence
c. (\bar{x}=12.5), (sigma = 1.2), (n = 50), 95 percent confidence

Explanation:

Step1: Find the critical value \(z_{\alpha/2}\)

For a 90% confidence interval, \(\alpha = 1 - 0.90=0.10\), so \(\alpha/2 = 0.05\). From the standard normal table, \(z_{0.05}\approx1.645\).
For a 99% confidence interval, \(\alpha = 1 - 0.99 = 0.01\), so \(\alpha/2=0.005\). From the standard normal table, \(z_{0.005}\approx2.576\).
For a 95% confidence interval, \(\alpha=1 - 0.95 = 0.05\), so \(\alpha/2 = 0.025\). From the standard normal table, \(z_{0.025}\approx1.96\).

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

The formula for the margin of error when the population standard deviation \(\sigma\) is known is \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\).

• For part (a):

\(E=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=1.645\times\frac{3}{\sqrt{10}}\approx1.645\times0.9487\approx1.56\)
The confidence interval is \(\bar{x}-E<\mu <\bar{x} + E\). Substituting \(\bar{x} = 24\) and \(E\approx1.56\), we get \(24 - 1.56<\mu<24 + 1.56\), or \(22.44<\mu<25.56\).

• For part (b):

\(E=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.576\times\frac{8}{\sqrt{25}}=2.576\times\frac{8}{5}=2.576\times1.6 = 4.1216\)
The confidence interval is \(\bar{x}-E<\mu<\bar{x}+E\). Substituting \(\bar{x}=125\) and \(E = 4.1216\), we get \(125-4.1216<\mu<125 + 4.1216\), or \(120.88<\mu<129.12\).

• For part (c):

\(E=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=1.96\times\frac{1.2}{\sqrt{50}}\approx1.96\times0.1697\approx0.33\)
The confidence interval is \(\bar{x}-E<\mu<\bar{x}+E\). Substituting \(\bar{x}=12.5\) and \(E\approx0.33\), we get \(12.5- 0.33<\mu<12.5+0.33\), or \(12.17<\mu<12.83\).

Answer:

a. \(22.44<\mu<25.56\)
b. \(120.88<\mu<129.12\)
c. \(12.17<\mu<12.83\)