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.
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.
MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
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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:
\[
\boxed{25.6} < \mu < \boxed{38.4}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa218d9b0-33cd-49fb-8966-615774819802%2F2729d926-840f-4a0d-bfd2-577b9df3f4ff%2F6mk1biq_processed.png&w=3840&q=75)
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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