A simple random sample of 50 items from a population with \(\sigma=6\) resulted in a sample mean of 32 a. Provide a \(90 \%\) confidence interval for the population mean. b. Provide a \(95 \%\) confidence interval for the population mean. c. Provide a \(99 \%\) confidence interval for the population mean.

The exercise asks for 90%, 95%, and 99% confidence intervals for the population mean. We will find the confidence intervals for each of these values separately. ##Step 2: Determine the z-scores for the desired confidence levels##

For the given confidence levels, we have: - 90% level: \(z_{\frac{\alpha}{2}} = 1.645\) - 95% level: \(z_{\frac{\alpha}{2}} = 1.960\) - 99% level: \(z_{\frac{\alpha}{2}} = 2.576\) These z-scores can be found using a z-table or statistical software. ##Step 3: Calculate the confidence intervals for each confidence level##

We'll now calculate the confidence intervals for each level, using the formula we mentioned earlier. a. 90% confidence interval: \[ \bar{x} \pm z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}} = 32 \pm 1.645\frac{6}{\sqrt{50}} = (30.573, 33.427) \] b. 95% confidence interval: \[ \bar{x} \pm z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}} = 32 \pm 1.960\frac{6}{\sqrt{50}} = (30.342, 33.658) \] c. 99% confidence interval: \[ \bar{x} \pm z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}} = 32 \pm 2.576\frac{6}{\sqrt{50}} = (29.860, 34.140) \] Therefore, we have the following confidence intervals for the population mean: - 90% confidence interval: \((30.573, 33.427)\) - 95% confidence interval: \((30.342, 33.658)\) - 99% confidence interval: \((29.860, 34.140)\)