Question

Find a \((1-\alpha) 100 \%\) confidence interval for a population mean \(\mu\) for these values: a. \(\alpha=.01, n=38, \bar{x}=34, s^{2}=12\) b. \(\alpha=.10, n=65, \bar{x}=1049, s^{2}=51\) c. \(\alpha=.05, n=89, \bar{x}=66.3, s^{2}=2.48\)

Short answer

a. The 99% confidence interval for the population mean, μ, in case a is (32.61, 35.39). b. The 90% confidence interval for the population mean, μ, in case b is (1046.50, 1051.50). c. The 95% confidence interval for the population mean, μ, in case c is (65.97, 66.63).

Step-by-step solution

a. Case \(\alpha=.01, n=38, \bar{x}=34, s^{2}=12\)

First, we need to calculate the standard error by dividing the sample standard deviation, s, by the square root of the sample size, n: $$SE = \frac{s}{\sqrt{n}} =\frac{\sqrt{12}}{\sqrt{38}} \approx 0.564$$ Then, we find the critical value from the t-distribution table (or using statistical software): $$t_{\frac{\alpha}{2}, n-1} = t_{0.005, 37} \approx 2.840$$ Now, we can plug the values into the confidence interval formula: $$(34 - 2.840 \cdot 0.564, 34 + 2.840 \cdot 0.564) \approx (32.61, 35.39)$$ The \(99\%\) confidence interval for the population mean, \(\mu\), in case a is \((32.61, 35.39)\).

b. Case \(\alpha=.10, n=65, \bar{x}=1049, s^{2}=51\)

First, calculate the standard error: $$SE = \frac{s}{\sqrt{n}} = \frac{\sqrt{51}}{\sqrt{65}} \approx 0.890$$ Then, find the critical value: $$t_{\frac{\alpha}{2}, n-1} = t_{0.05, 64} \approx 1.669$$ Plug the values into the confidence interval formula: $$(1049 - 1.669 \cdot 0.890, 1049 + 1.669 \cdot 0.890) \approx (1046.50, 1051.50)$$ The \(90\%\) confidence interval for the population mean, \(\mu\), in case b is \((1046.50, 1051.50)\).

c. Case \(\alpha=.05, n=89, \bar{x}=66.3, s^{2}=2.48\)

First, calculate the standard error: $$SE = \frac{s}{\sqrt{n}} = \frac{\sqrt{2.48}}{\sqrt{89}} \approx 0.166$$ Then, find the critical value: $$t_{\frac{\alpha}{2}, n-1} = t_{0.025, 88} \approx 1.989$$ Plug the values into the confidence interval formula: $$(66.3 - 1.989 \cdot 0.166, 66.3 + 1.989 \cdot 0.166) \approx (65.97, 66.63)$$ The \(95\%\) confidence interval for the population mean, \(\mu\), in case c is \((65.97, 66.63)\).