In a survey, 18 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $37 and standard deviation of $10. Construct a confidence interval at a 99% confidence level. Give your answers to one decimal place. ± Interpret your confidence interval in the context of this problem.
In a survey, 18 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $37 and standard deviation of $10. Construct a confidence interval at a 99% confidence level. Give your answers to one decimal place. ± Interpret your confidence interval in the context of this problem.
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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![### Construction of a Confidence Interval
**Problem Description:**
In a survey, 18 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $37 and a standard deviation of $10. Construct a confidence interval at a 99% confidence level.
**Instructions:**
Give your answers to one decimal place.
__________ ± __________
**Interpretation:**
Interpret your confidence interval in the context of this problem.
**Detailed Explanation:**
To construct the confidence interval for the given data:
1. **Sample Mean (M)**: $37
2. **Standard Deviation (SD)**: $10
3. **Sample Size (n)**: 18
4. **Confidence Level**: 99%
For a confidence interval at 99%, we use the Z-distribution (assuming a normal distribution due to a roughly bell-shaped data set). The critical value (Z-score) for a 99% confidence level is approximately 2.576.
The standard error (SE) is calculated as:
\[
SE = \frac{SD}{\sqrt{n}} = \frac{10}{\sqrt{18}} \approx 2.4
\]
Next, calculate the margin of error (ME):
\[
ME = Z \times SE = 2.576 \times 2.4 \approx 6.2
\]
To construct the confidence interval, add and subtract the margin of error from the sample mean:
\[
CI = M \pm ME = 37 \pm 6.2
\]
Thus, the confidence interval is:
\[
30.8 \leq \mu \leq 43.2
\]
**Interpretation:**
The 99% confidence interval means that we are 99% confident that the true mean amount spent on a child’s last birthday gift falls between $30.8 and $43.2.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc7e5ddd4-9774-49f5-a3a5-694b08ae0d81%2F2faf2954-f9e6-4a71-9b65-5615b2abbffa%2Fbhsrwfp_processed.png&w=3840&q=75)
Transcribed Image Text:### Construction of a Confidence Interval
**Problem Description:**
In a survey, 18 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $37 and a standard deviation of $10. Construct a confidence interval at a 99% confidence level.
**Instructions:**
Give your answers to one decimal place.
__________ ± __________
**Interpretation:**
Interpret your confidence interval in the context of this problem.
**Detailed Explanation:**
To construct the confidence interval for the given data:
1. **Sample Mean (M)**: $37
2. **Standard Deviation (SD)**: $10
3. **Sample Size (n)**: 18
4. **Confidence Level**: 99%
For a confidence interval at 99%, we use the Z-distribution (assuming a normal distribution due to a roughly bell-shaped data set). The critical value (Z-score) for a 99% confidence level is approximately 2.576.
The standard error (SE) is calculated as:
\[
SE = \frac{SD}{\sqrt{n}} = \frac{10}{\sqrt{18}} \approx 2.4
\]
Next, calculate the margin of error (ME):
\[
ME = Z \times SE = 2.576 \times 2.4 \approx 6.2
\]
To construct the confidence interval, add and subtract the margin of error from the sample mean:
\[
CI = M \pm ME = 37 \pm 6.2
\]
Thus, the confidence interval is:
\[
30.8 \leq \mu \leq 43.2
\]
**Interpretation:**
The 99% confidence interval means that we are 99% confident that the true mean amount spent on a child’s last birthday gift falls between $30.8 and $43.2.
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