To construct a 98% confidence interval for the mean level of carbon monoxide using the given measurements, we need to follow these steps:

Given Data:

• Measurements (in picocuries per liter):
  $314.3, 348.2, 371, 289.3, 288.5, 338.8$

1. Step 1: Calculate the Sample Mean ($\bar{x}$)

$\bar{x} = \frac{314.3 + 348.2 + 371 + 289.3 + 288.5 + 338.8}{6}$ $\bar{x} = \frac{1950.1}{6} \approx 325.02$

2. Step 2: Calculate the Sample Standard Deviation (s)

The sample standard deviation formula is:

$s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$

Let's calculate the squared deviations and their sum:

(314.3 - 325.02)^2 & = 114.22 \\ (348.2 - 325.02)^2 & = 538.71 \\ (371 - 325.02)^2 & = 2119.08 \\ (289.3 - 325.02)^2 & = 1278.21 \\ (288.5 - 325.02)^2 & = 1337.14 \\ (338.8 - 325.02)^2 & = 188.07 \\ \end{aligned}$$ Sum of squared deviations: $$114.22 + 538.71 + 2119.08 + 1278.21 + 1337.14 + 188.07 = 5575.43$$ Now calculate $$s$$: $$s = \sqrt{\frac{5575.43}{6 - 1}} = \sqrt{\frac{5575.43}{5}} \approx \sqrt{1115.09} \approx 33.41$$ 3. **Step 3: Determine the t-value** Since the sample size is $$n = 6$$, degrees of freedom $$df = 6 - 1 = 5$$. For a 98% confidence interval and $$df = 5$$, the critical t-value ($$t_{\alpha/2}$$) is approximately 2.571 (based on t-distribution tables). 4. **Step 4: Calculate the Margin of Error (E)** $$E = t_{\alpha/2} \times \frac{s}{\sqrt{n}}$$ Substitute the values: $$E = 2.571 \times \frac{33.41}{\sqrt{6}} = 2.571 \times \frac{33.41}{2.449} \approx 2.571 \times 13.64 \approx 35.07$$ 5. **Step 5: Calculate the Confidence Interval** The confidence interval is given by: $$\left( \bar{x} - E, \bar{x} + E \right)$$ Substitute the values: $$\left( 325.02 - 35.07, 325.02 + 35.07 \right)$$ $$\left( 289.95, 360.09 \right)$$ ### Final Answer: The 98% confidence interval for the mean level of carbon monoxide is approximately **(289.95, 360.09)** picocuries per liter. Would you like more details or have any questions? Here are 5 related questions to deepen your understanding: 1. What assumptions are required to construct a confidence interval using the t-distribution? 2. How does the confidence level affect the width of the confidence interval? 3. What happens to the margin of error if the sample size increases? 4. Why is the t-distribution used instead of the z-distribution for small samples? 5. How would the confidence interval change if we wanted a 95% confidence level instead of 98%? **Tip:** Larger sample sizes generally lead to more precise confidence intervals, as they reduce the margin of error.