27. In a random sample of six mobile devices, the mean repair cost was $65.00 and the standard dev $14.00. Assume the population is normally distributed and use a t-distribution to find the margin o construct a 90% confidence interval for the population mean. Interpret the results.
27. In a random sample of six mobile devices, the mean repair cost was $65.00 and the standard dev $14.00. Assume the population is normally distributed and use a t-distribution to find the margin o construct a 90% confidence interval for the population mean. Interpret the results.
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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![**Problem 27: Confidence Interval Calculation for Mobile Device Repair Costs**
In a random sample of six mobile devices, the mean repair cost was $65.00 and the standard deviation was $14.00. Assume the population is normally distributed, and use a t-distribution to find the margin of error and construct a 90% confidence interval for the population mean. Interpret the results.
**Solution Steps:**
1. **Identify the given information:**
- Sample size (n): 6
- Sample mean (\(\bar{x}\)): $65.00
- Sample standard deviation (s): $14.00
- Confidence level: 90%
2. **Determine the appropriate t-score:**
- Degrees of freedom (df) = n - 1 = 6 - 1 = 5
- For a 90% confidence level and 5 degrees of freedom, use a t-table or calculator to find the critical t-value (t*).
3. **Calculate the margin of error (E):**
\[
E = t^* \times \left(\frac{s}{\sqrt{n}}\right)
\]
4. **Construct the confidence interval:**
\[
\text{Lower bound} = \bar{x} - E
\]
\[
\text{Upper bound} = \bar{x} + E
\]
5. **Interpret the results:**
- The interval will provide the range in which we are 90% confident the true mean repair cost for the population of mobile devices falls.
**Detailed Steps:**
1. Given:
- \(\bar{x} = 65\)
- s = 14
- n = 6
- Confidence level = 90%
2. Degrees of freedom (df) = 6 - 1 = 5
Using a t-table or calculator for a 90% confidence level with df = 5, the critical t-value (t*) is approximately 2.015.
3. Margin of error (E):
\[
E = 2.015 \times \left(\frac{14}{\sqrt{6}}\right)
\]
\[
E = 2.015 \times \left(\frac{14}{2.4495}\right)
\]
\[
E = 2.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9c9e2dde-1c19-44ed-a9d3-170191d095c3%2Fea3064ef-1236-497c-89c9-35de9eb11f52%2Fuce9gie_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 27: Confidence Interval Calculation for Mobile Device Repair Costs**
In a random sample of six mobile devices, the mean repair cost was $65.00 and the standard deviation was $14.00. Assume the population is normally distributed, and use a t-distribution to find the margin of error and construct a 90% confidence interval for the population mean. Interpret the results.
**Solution Steps:**
1. **Identify the given information:**
- Sample size (n): 6
- Sample mean (\(\bar{x}\)): $65.00
- Sample standard deviation (s): $14.00
- Confidence level: 90%
2. **Determine the appropriate t-score:**
- Degrees of freedom (df) = n - 1 = 6 - 1 = 5
- For a 90% confidence level and 5 degrees of freedom, use a t-table or calculator to find the critical t-value (t*).
3. **Calculate the margin of error (E):**
\[
E = t^* \times \left(\frac{s}{\sqrt{n}}\right)
\]
4. **Construct the confidence interval:**
\[
\text{Lower bound} = \bar{x} - E
\]
\[
\text{Upper bound} = \bar{x} + E
\]
5. **Interpret the results:**
- The interval will provide the range in which we are 90% confident the true mean repair cost for the population of mobile devices falls.
**Detailed Steps:**
1. Given:
- \(\bar{x} = 65\)
- s = 14
- n = 6
- Confidence level = 90%
2. Degrees of freedom (df) = 6 - 1 = 5
Using a t-table or calculator for a 90% confidence level with df = 5, the critical t-value (t*) is approximately 2.015.
3. Margin of error (E):
\[
E = 2.015 \times \left(\frac{14}{\sqrt{6}}\right)
\]
\[
E = 2.015 \times \left(\frac{14}{2.4495}\right)
\]
\[
E = 2.
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