3. Walter usually meets Julia at the track. He prefers to jog 3 miles. From long experience, he knows that the standard deviation for his jogging times is 2.40 minutes. For a random sample of 90 jogging sessions, the mean time was 2.25 minutes. Find a 99% confidence interval for the mean jogging time for all of Walter's 3-mile running times over the past several years.
3. Walter usually meets Julia at the track. He prefers to jog 3 miles. From long experience, he knows that the standard deviation for his jogging times is 2.40 minutes. For a random sample of 90 jogging sessions, the mean time was 2.25 minutes. Find a 99% confidence interval for the mean jogging time for all of Walter's 3-mile running times over the past several years.
MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
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Chapter1: Starting With Matlab
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![**Problem Statement:**
3. Walter usually meets Julia at the track. He prefers to jog 3 miles. From long experience, he knows that the standard deviation for his jogging times is 2.40 minutes. For a random sample of 90 jogging sessions, the mean time was 22.5 minutes. Find a 99% confidence interval for the mean jogging time for all of Walter's 3-mile running times over the past several years.
**Solution Approach:**
To calculate the 99% confidence interval for the mean jogging time, use the following formula for the confidence interval of the mean:
\[ \text{Confidence Interval} = \bar{x} \pm z \left(\frac{\sigma}{\sqrt{n}}\right) \]
Where:
- \(\bar{x}\) is the sample mean (22.5 minutes).
- \(z\) is the z-score corresponding to the confidence level (99%). For 99%, \(z \approx 2.576\).
- \(\sigma\) is the population standard deviation (2.40 minutes).
- \(n\) is the sample size (90).
**Steps:**
1. Calculate the standard error (SE):
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{2.40}{\sqrt{90}} \]
2. Multiply the standard error by the z-score:
\[ ME = 2.576 \times SE \]
(ME is the margin of error)
3. Determine the confidence interval:
\[ \text{Lower limit} = \bar{x} - ME \]
\[ \text{Upper limit} = \bar{x} + ME \]
These steps will provide the 99% confidence interval for the mean jogging time for all of Walter's 3-mile runs over the past several years.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcb64ed77-0e71-4b8d-9699-28308f43c665%2F36121baa-be78-4cd6-bd85-306aab22a927%2Ftwb7joe_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
3. Walter usually meets Julia at the track. He prefers to jog 3 miles. From long experience, he knows that the standard deviation for his jogging times is 2.40 minutes. For a random sample of 90 jogging sessions, the mean time was 22.5 minutes. Find a 99% confidence interval for the mean jogging time for all of Walter's 3-mile running times over the past several years.
**Solution Approach:**
To calculate the 99% confidence interval for the mean jogging time, use the following formula for the confidence interval of the mean:
\[ \text{Confidence Interval} = \bar{x} \pm z \left(\frac{\sigma}{\sqrt{n}}\right) \]
Where:
- \(\bar{x}\) is the sample mean (22.5 minutes).
- \(z\) is the z-score corresponding to the confidence level (99%). For 99%, \(z \approx 2.576\).
- \(\sigma\) is the population standard deviation (2.40 minutes).
- \(n\) is the sample size (90).
**Steps:**
1. Calculate the standard error (SE):
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{2.40}{\sqrt{90}} \]
2. Multiply the standard error by the z-score:
\[ ME = 2.576 \times SE \]
(ME is the margin of error)
3. Determine the confidence interval:
\[ \text{Lower limit} = \bar{x} - ME \]
\[ \text{Upper limit} = \bar{x} + ME \]
These steps will provide the 99% confidence interval for the mean jogging time for all of Walter's 3-mile runs over the past several years.
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