Question 2 of 10 , Step 2 of 2 In a random sample of 7 residents of the state of Montana, the mean waste recycled-per person per day was 2.8 pounds with a standard deviation of 0.16 pounds. Determine the \( 90 \% \) confidence interval for the mean waste recycled per person per day for the population of Montana. Assume the population is approximately normal. Step 2 of 2 : Construct the \( 90 \% \) confidence interval. Round your answer to one decimal place.

To calculate the 90% confidence interval for the mean waste recycled per person per day in Montana, you can use the formula: \[ \text{Confidence Interval} = \bar{x} \pm t_{\alpha/2} \cdot \left(\frac{s}{\sqrt{n}}\right) \] Here, - \(\bar{x}\) is the sample mean, which is 2.8 pounds. - \(s\) is the sample standard deviation, which is 0.16 pounds. - \(n\) is the sample size, which is 7. - \(t_{\alpha/2}\) is the t-value that corresponds to 90% confidence and degrees of freedom (df = n-1 = 6). Using a t-table, for \(df = 6\) and a 90% confidence level, \(t_{\alpha/2} \approx 1.943\). Now, calculate the standard error: \[ \text{Standard Error} = \frac{s}{\sqrt{n}} = \frac{0.16}{\sqrt{7}} \approx 0.0605 \] Now, we can compute the margin of error: \[ \text{Margin of Error} = t_{\alpha/2} \cdot \text{Standard Error} \approx 1.943 \cdot 0.0605 \approx 0.1177 \] Finally, construct the confidence interval: \[ \text{Lower limit} = 2.8 - 0.1177 \approx 2.7 \] \[ \text{Upper limit} = 2.8 + 0.1177 \approx 2.9 \] Therefore, the 90% confidence interval for the mean waste recycled per person per day for the population of Montana is approximately \((2.7, 2.9)\) pounds.