A group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2 on a placement test. What is the 99% confidence interval for the mean score,, of all students taking the test? Round to one decimal place. <μ<
A group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2 on a placement test. What is the 99% confidence interval for the mean score,, of all students taking the test? Round to one decimal place. <μ<
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:**
A group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2 on a placement test.
What is the 99% confidence interval for the mean score, µ, of all students taking the test? *Round to one decimal place.*
**Answer:**
[ ] < µ < [ ]
**Explanation:**
To solve for the 99% confidence interval, use the following steps:
1. Determine the sample mean (x̄), which is 29.5.
2. Find the standard deviation (σ), which is 5.2.
3. Determine the sample size (n), which is 59.
4. Find the critical value (z) for a 99% confidence interval. For the 99% confidence level, the z-score is approximately 2.576.
5. Calculate the standard error (SE) of the mean using the formula:
SE = σ / √n
SE = 5.2 / √59 ≈ 0.6762
6. Calculate the margin of error (ME) using the formula:
ME = z * SE
ME ≈ 2.576 * 0.6762 ≈ 1.7421
7. Construct the confidence interval using the formula:
CI = x̄ ± ME
CI = 29.5 ± 1.7421
Therefore, the 99% confidence interval for the mean score is approximately:
27.8 < µ < 31.3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fed8faccd-6cf6-4e70-b90c-dd265de0d7ab%2F05a4a7a8-fc93-4fb2-9e6c-ff84aa05f5aa%2Fe8ayxhb_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
A group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2 on a placement test.
What is the 99% confidence interval for the mean score, µ, of all students taking the test? *Round to one decimal place.*
**Answer:**
[ ] < µ < [ ]
**Explanation:**
To solve for the 99% confidence interval, use the following steps:
1. Determine the sample mean (x̄), which is 29.5.
2. Find the standard deviation (σ), which is 5.2.
3. Determine the sample size (n), which is 59.
4. Find the critical value (z) for a 99% confidence interval. For the 99% confidence level, the z-score is approximately 2.576.
5. Calculate the standard error (SE) of the mean using the formula:
SE = σ / √n
SE = 5.2 / √59 ≈ 0.6762
6. Calculate the margin of error (ME) using the formula:
ME = z * SE
ME ≈ 2.576 * 0.6762 ≈ 1.7421
7. Construct the confidence interval using the formula:
CI = x̄ ± ME
CI = 29.5 ± 1.7421
Therefore, the 99% confidence interval for the mean score is approximately:
27.8 < µ < 31.3
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