The following data represent the concentration of dissolved organic carbon​ (mg/L) collected from 20 samples of organic soil. Assume that the population is normally distributed. Complete parts​ (a) through​ (c) on the right.

14.0014.00

29.80

27.10

16.51

16.8716.87

8.81

14.0014.00

20.46

14.90

33.67

30.91

14.86

10.3010.30

15.35

9.72

19.80

14.86

8.09

20.4620.46

18.30

Question content area bottom

Part 1

​(a) Find the sample mean.

The sample mean is

17.9417.94.

​(Round to two decimal places as​ needed.)

Part 2

​(b) Find the sample standard deviation.

The sample standard deviation is

7.377.37.

​(Round to two decimal places as​ needed.)

Part 3

​(c)**** Construct a

9595​%

confidence interval for the population mean

muμ.

The

9595​%

confidence interval for the population mean

muμ

is

​(enter your response here​,enter your response here​,).

​(Round to two decimal places as​ needed.)

To construct a 95% confidence interval for the population mean \( \mu \), we can follow the steps outlined below:

### Step 1: Identify the Given Information
- Sample mean \( \bar{x} = 17.94 \)
- Sample standard deviation \( s = 7.37 \)
- Sample size \( n = 20 \)
- Confidence level = 95%, so the significance level \( \alpha = 0.05 \), and the critical value \( t_{\alpha/2} \) is based on the t-distribution with \( n - 1 = 19 \) degrees of freedom.

### Step 2: Find the Critical Value
For a 95% confidence level with 19 degrees of freedom, we need the critical value \( t_{\alpha/2} \) from the t-distribution table.

From standard tables, the critical value \( t_{\alpha/2} \) for \( \alpha = 0.05 \) and 19 degrees of freedom is approximately **2.093**.

### Step 3: Calculate the Margin of Error (ME)
The margin of error is given by the formula:

\[
\text{ME} = t_{\alpha/2} \times \frac{s}{\sqrt{n}}
\]

Substituting the values:

\[
\text{ME} = 2.093 \times \frac{7.37}{\sqrt{20}} = 2.093 \times 1.648 = 3.448
\]

### Step 4: Construct the Confidence Interval
The confidence interval is calculated as:

\[
\text{CI} = \bar{x} \pm \text{ME}
\]

Substitute the values:

\[
\text{CI} = 17.94 \pm 3.45
\]

So the confidence interval is:

\[
(17.94 - 3.45, 17.94 + 3.45) = (14.49, 21.39)
\]

### Conclusion
The 95% confidence interval for the population mean \( \mu \) is **(14.49, 21.39)**.

Would you like more details on the t-distribution or any other part of this calculation?

---

Here are five related questions to expand your understanding:
1. How would the confidence interval change if the sample size were larger?
2. What is the effect of increasing the confidence level (e.g., from 95% to 99%) on the interval?
3. Why do we use the t-distribution instead of the z-distribution for small sample sizes?
4. How is the critical value of the t-distribution calculated?
5. What assumptions must be met to use the t-distribution for this confidence interval?

**Tip:** Increasing the sample size decreases the margin of error, leading to a narrower confidence interval for the mean.

The following data represent the concentration of dissolved organic carbon​ (mg/L) collected from 20 samples of organic soil. Assume that the population is normally distributed. Complete parts​ (a) through​ (c): (a) Find the sample mean, (b) Find the sample standard deviation, and (c) Construct a 95% confidence interval for the population mean μ.

Learn how to calculate the sample mean, standard deviation, and 95% confidence interval for dissolved organic carbon concentration from 20 soil samples. This statistical problem involves the use of t-distribution and is suitable for college-level statistics students.

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