The following measurements (in picocuries per liter) were recorded by a set of carbon monoxide detectors installed in a laboratory facility:  

314.3,348.2,371,289.3,288.5,338.8

Using these measurements, construct a 98%
 confidence interval for the mean level of carbon monoxide present in the facility.  Assume the population is approximately normal.


Step 4 of 4 :  Construct the 98%
 confidence interval. Round your answer to two decimal places.

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:

\[
\begin{aligned}
(314.3 - 325.02)^2 & = 114.22 \\
(348.2 - 325.02)^2 & = 538.71 \\
(371 - 325.02)^2 & = 2119.08 \\
(289.3 - 325.02)^2 & = 1278.21 \\
(288.5 - 325.02)^2 & = 1337.14 \\
(338.8 - 325.02)^2 & = 188.07 \\
\end{aligned}
\]

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.

The following measurements (in picocuries per liter) were recorded by a set of carbon monoxide detectors installed in a laboratory facility: 314.3, 348.2, 371, 289.3, 288.5, 338.8. Using these measurements, construct a 98% confidence interval for the mean level of carbon monoxide present in the facility. Assume the population is approximately normal.

Construct a 98% confidence interval for the mean level of carbon monoxide detected in a laboratory using sample measurements. This problem covers statistical concepts, including sample mean, standard deviation, margin of error, and t-distribution. Learn how to apply these formulas to real-world data.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation