Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 415 drivers and find that 300 claim to always buckle up. Construct a 96% confidence interval for the population proportion that claim to always buckle up.
Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 415 drivers and find that 300 claim to always buckle up. Construct a 96% confidence interval for the population proportion that claim to always buckle up.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![**Question 6**
Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 415 drivers and find that 300 claim to always buckle up. Construct a 96% confidence interval for the population proportion that claim to always buckle up.
---
### Explanation
To construct a 96% confidence interval for the population proportion, follow these steps:
1. **Find the sample proportion (p̂):**
\[
p̂ = \frac{\text{Number of drivers who buckle up}}{\text{Total number of drivers surveyed}} = \frac{300}{415}
\]
2. **Determine the standard error (SE):**
\[
SE = \sqrt{\frac{p̂(1 - p̂)}{n}}
\]
where \( n = 415 \).
3. **Find the Z-score for a 96% confidence interval:**
The Z-score corresponds to the middle 96% of the standard normal distribution.
4. **Calculate the confidence interval:**
\[
\text{Confidence interval} = p̂ \pm (Z \times SE)
\]
This will provide the range in which the true population proportion is likely to fall, with 96% confidence.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feb2e15ef-8640-49b0-a75f-b139085238e1%2F74404ee7-d1ce-423a-8c41-0d9fe83e0823%2Fnzgktfn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 6**
Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 415 drivers and find that 300 claim to always buckle up. Construct a 96% confidence interval for the population proportion that claim to always buckle up.
---
### Explanation
To construct a 96% confidence interval for the population proportion, follow these steps:
1. **Find the sample proportion (p̂):**
\[
p̂ = \frac{\text{Number of drivers who buckle up}}{\text{Total number of drivers surveyed}} = \frac{300}{415}
\]
2. **Determine the standard error (SE):**
\[
SE = \sqrt{\frac{p̂(1 - p̂)}{n}}
\]
where \( n = 415 \).
3. **Find the Z-score for a 96% confidence interval:**
The Z-score corresponds to the middle 96% of the standard normal distribution.
4. **Calculate the confidence interval:**
\[
\text{Confidence interval} = p̂ \pm (Z \times SE)
\]
This will provide the range in which the true population proportion is likely to fall, with 96% confidence.
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