If n=19, (x-bar)=32, and s=11, construct a confidence interval at a 98% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place.

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### Constructing a 98% Confidence Interval

Given the following data:
- Sample size (\(n\)) = 19
- Sample mean (\(\overline{x}\)) = 32
- Sample standard deviation (\(s\)) = 11

We are tasked with constructing a confidence interval at a 98% confidence level. Assume the data came from a normally distributed population.

#### Steps to Construct the Confidence Interval

1. **Identify the sample size (\(n\)), sample mean (\(\overline{x}\)), and sample standard deviation (\(s\))**:
   - \(n = 19\)
   - \(\overline{x} = 32\)
   - \(s = 11\)

2. **Determine the appropriate t-score for a 98% confidence level**:
   - Degrees of freedom (\(df\)) = \(n - 1 = 19 - 1 = 18\)
   - Using a t-distribution table or calculator, find the t-score for 98% confidence level with \(df = 18\). The value is approximately \(t \approx 2.552\).

3. **Calculate the standard error of the mean (SEM)**:
   \[
   \text{SEM} = \frac{s}{\sqrt{n}} = \frac{11}{\sqrt{19}} \approx 2.524
   \]

4. **Compute the margin of error (ME)**:
   \[
   \text{ME} = t \times \text{SEM} = 2.552 \times 2.524 \approx 6.4
   \]

5. **Determine the confidence interval**:
   - Lower limit: \(\overline{x} - \text{ME} = 32 - 6.4 = 25.6\)
   - Upper limit: \(\overline{x} + \text{ME} = 32 + 6.4 = 38.4\)

So, the 98% confidence interval for the population mean is:

\[
25.6 < \mu < 38.4
\]

#### Summary:

Fill in the results in the boxes provided:
\[
\boxed{25.6} < \mu < \boxed{38.4}
\]
Transcribed Image Text:### Constructing a 98% Confidence Interval Given the following data: - Sample size (\(n\)) = 19 - Sample mean (\(\overline{x}\)) = 32 - Sample standard deviation (\(s\)) = 11 We are tasked with constructing a confidence interval at a 98% confidence level. Assume the data came from a normally distributed population. #### Steps to Construct the Confidence Interval 1. **Identify the sample size (\(n\)), sample mean (\(\overline{x}\)), and sample standard deviation (\(s\))**: - \(n = 19\) - \(\overline{x} = 32\) - \(s = 11\) 2. **Determine the appropriate t-score for a 98% confidence level**: - Degrees of freedom (\(df\)) = \(n - 1 = 19 - 1 = 18\) - Using a t-distribution table or calculator, find the t-score for 98% confidence level with \(df = 18\). The value is approximately \(t \approx 2.552\). 3. **Calculate the standard error of the mean (SEM)**: \[ \text{SEM} = \frac{s}{\sqrt{n}} = \frac{11}{\sqrt{19}} \approx 2.524 \] 4. **Compute the margin of error (ME)**: \[ \text{ME} = t \times \text{SEM} = 2.552 \times 2.524 \approx 6.4 \] 5. **Determine the confidence interval**: - Lower limit: \(\overline{x} - \text{ME} = 32 - 6.4 = 25.6\) - Upper limit: \(\overline{x} + \text{ME} = 32 + 6.4 = 38.4\) So, the 98% confidence interval for the population mean is: \[ 25.6 < \mu < 38.4 \] #### Summary: Fill in the results in the boxes provided: \[ \boxed{25.6} < \mu < \boxed{38.4} \]
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