MATLAB: An Introduction with Applications
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ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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![### Educational Television
In a random sample of 190 people, 145 said that they watched educational television. Find the 99% confidence interval of the true proportion of people who watched educational television. Round the answers to at least three decimal places.
\[
\boxed{} \quad <p< \quad \boxed{}
\]
**Instructions and Explanation:**
To calculate the 99% confidence interval for the true proportion of people who watched educational television, we follow these steps:
1. **Calculate the sample proportion (\(\hat{p}\))**:
\[
\hat{p} = \frac{x}{n} = \frac{145}{190}
\]
2. **Determine the standard error (SE) of the proportion**:
\[
SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}
\]
3. **Find the Z-value for a 99% confidence level** (from a standard normal table, the Z-value is approximately 2.576 for a two-tailed test):
4. **Compute the margin of error (ME)**:
\[
ME = Z \times SE
\]
5. **Construct the confidence interval**:
\[
\hat{p} - ME < p < \hat{p} + ME
\]
Use this method to fill in the provided answer boxes, rounding your final answers to at least three decimal places. This process will allow you to determine the range within which the true proportion of people who watch educational television is likely to fall, with 99% confidence.](https://content.bartleby.com/qna-images/question/176f32ac-c9e1-4f00-8dba-82b002b2f053/6978131b-7ba9-45d4-8bc9-71b139d4fdbe/ka2px9q_processed.jpeg)
Transcribed Image Text:### Educational Television
In a random sample of 190 people, 145 said that they watched educational television. Find the 99% confidence interval of the true proportion of people who watched educational television. Round the answers to at least three decimal places.
\[
\boxed{} \quad <p< \quad \boxed{}
\]
**Instructions and Explanation:**
To calculate the 99% confidence interval for the true proportion of people who watched educational television, we follow these steps:
1. **Calculate the sample proportion (\(\hat{p}\))**:
\[
\hat{p} = \frac{x}{n} = \frac{145}{190}
\]
2. **Determine the standard error (SE) of the proportion**:
\[
SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}
\]
3. **Find the Z-value for a 99% confidence level** (from a standard normal table, the Z-value is approximately 2.576 for a two-tailed test):
4. **Compute the margin of error (ME)**:
\[
ME = Z \times SE
\]
5. **Construct the confidence interval**:
\[
\hat{p} - ME < p < \hat{p} + ME
\]
Use this method to fill in the provided answer boxes, rounding your final answers to at least three decimal places. This process will allow you to determine the range within which the true proportion of people who watch educational television is likely to fall, with 99% confidence.
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