Questions: If n=16, x̄=32, and s=4, construct a confidence interval at a 98% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place. <μ<

Solution Steps

We are provided with the following parameters for constructing a confidence interval for the population mean:

• Sample size: $n = 16$
• Sample mean: $\bar{x} = 32$
• Sample standard deviation: $s = 4$
• Confidence level: $98\%$

The significance level ($\alpha$) is calculated as: $$\alpha = 1 - \text{Confidence Level} = 1 - 0.98 = 0.02$$

To calculate the margin of error, we use the formula: $$\text{Margin of Error} = t \cdot \frac{s}{\sqrt{n}}$$ where $t$ is the critical value from the t-distribution for $n - 1 = 15$ degrees of freedom at the $98\%$ confidence level. From the calculations, we find: $$t \approx 2.6$$ Thus, the margin of error is: $$\text{Margin of Error} = 2.6 \cdot \frac{4}{\sqrt{16}} = 2.6 \cdot \frac{4}{4} = 2.6$$

The confidence interval for the population mean is given by: $$\bar{x} \pm \text{Margin of Error}$$ Substituting the values, we have: $$32 \pm 2.6$$ This results in the interval: $$(32 - 2.6, 32 + 2.6) = (29.4, 34.6)$$

Final Answer