Construct the indicated confidence interval for the population mean μ using the t-distribution. Assume the population is normally distributed. c=0.95, x=13.4, s=3.0, n=8 (Round to one decimal place as needed.) G
Construct the indicated confidence interval for the population mean μ using the t-distribution. Assume the population is normally distributed. c=0.95, x=13.4, s=3.0, n=8 (Round to one decimal place as needed.) G
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![**Constructing Confidence Intervals Using the t-Distribution**
In this example, we aim to construct the indicated confidence interval for the population mean \( \mu \) using the t-distribution. The assumption provided is that the population is normally distributed. Below are the specific details provided:
- Confidence level (\( c \)): 0.95
- Sample mean (\( \overline{x} \)): 13.4
- Sample standard deviation (\( s \)): 3.0
- Sample size (\( n \)): 8
### Calculating the Confidence Interval
To find the confidence interval, follow these steps:
1. **Determine the degrees of freedom (df):**
\[
\text{df} = n - 1 = 8 - 1 = 7
\]
2. **Find the t-value for the given confidence level and degrees of freedom:**
Use a t-distribution table or a statistical software to find the t-value that corresponds to a 95% confidence level with 7 degrees of freedom.
3. **Calculate the margin of error (E):**
\[
E = t_{\alpha/2} \times \frac{s}{\sqrt{n}}
\]
where \( t_{\alpha/2} \) is the t-value from the t-distribution table.
4. **Construct the confidence interval:**
\[
\left( \overline{x} - E, \overline{x} + E \right)
\]
### Example Illustration
Let’s assume we've found the t-value for a 95% confidence level and 7 degrees of freedom (df) to be approximately 2.365.
1. **Calculating the margin of error (E):**
\[
E = 2.365 \times \frac{3.0}{\sqrt{8}} \approx 2.5
\]
2. **Construct the confidence interval:**
\[
\left( 13.4 - 2.5, 13.4 + 2.5 \right) = \left( 10.9, 15.9 \right)
\]
### Final Confidence Interval
\[
\left( 10.9, 15.9 \right)
\]
### Note:
*Round figures to one decimal place as needed.*
Thus, we can be 95](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1a5e68af-43fc-4f18-979d-78274e50dd4a%2Ff07dda23-5c04-4f1d-a0c9-6fa47590e2e6%2Fxvxkrl_processed.png&w=3840&q=75)
Transcribed Image Text:**Constructing Confidence Intervals Using the t-Distribution**
In this example, we aim to construct the indicated confidence interval for the population mean \( \mu \) using the t-distribution. The assumption provided is that the population is normally distributed. Below are the specific details provided:
- Confidence level (\( c \)): 0.95
- Sample mean (\( \overline{x} \)): 13.4
- Sample standard deviation (\( s \)): 3.0
- Sample size (\( n \)): 8
### Calculating the Confidence Interval
To find the confidence interval, follow these steps:
1. **Determine the degrees of freedom (df):**
\[
\text{df} = n - 1 = 8 - 1 = 7
\]
2. **Find the t-value for the given confidence level and degrees of freedom:**
Use a t-distribution table or a statistical software to find the t-value that corresponds to a 95% confidence level with 7 degrees of freedom.
3. **Calculate the margin of error (E):**
\[
E = t_{\alpha/2} \times \frac{s}{\sqrt{n}}
\]
where \( t_{\alpha/2} \) is the t-value from the t-distribution table.
4. **Construct the confidence interval:**
\[
\left( \overline{x} - E, \overline{x} + E \right)
\]
### Example Illustration
Let’s assume we've found the t-value for a 95% confidence level and 7 degrees of freedom (df) to be approximately 2.365.
1. **Calculating the margin of error (E):**
\[
E = 2.365 \times \frac{3.0}{\sqrt{8}} \approx 2.5
\]
2. **Construct the confidence interval:**
\[
\left( 13.4 - 2.5, 13.4 + 2.5 \right) = \left( 10.9, 15.9 \right)
\]
### Final Confidence Interval
\[
\left( 10.9, 15.9 \right)
\]
### Note:
*Round figures to one decimal place as needed.*
Thus, we can be 95
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