In a random sample of four mobile​ devices, the mean repair cost was ​$85.00 and the standard deviation was ​$12.50. 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.
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Part 1
The 90​% confidence interval for the population mean mu is ​(
  
enter your response here​,
  
enter your response here​). 
​(Round to two decimal places as​ needed.)

To construct the 90% confidence interval for the population mean $ \mu $ and find the margin of error, we can use the formula for the confidence interval for the mean when the population standard deviation is unknown and a $ t $-distribution is appropriate.

### Step 1: Calculate the Margin of Error
The formula for the margin of error $ E $ using a $ t $-distribution is:
\[
E = t \times \frac{s}{\sqrt{n}}
\]
where:
- $ t $ is the critical value from the $ t $-distribution for a 90% confidence level with $ n-1 $ degrees of freedom.
- $ s $ is the sample standard deviation, which is 12.50.
- $ n $ is the sample size, which is 4.

For a 90% confidence level with $ n-1 = 3 $ degrees of freedom, the critical $ t $-value (using $ t_{0.05,3} $) is approximately 2.353.

Now we can substitute in the values:
\[
E = 2.353 \times \frac{12.50}{\sqrt{4}}
\]
\[
E = 2.353 \times \frac{12.50}{2}
\]
\[
E = 2.353 \times 6.25
\]
\[
E = 14.71
\]

### Step 2: Construct the Confidence Interval
The confidence interval for the population mean $ \mu $ is given by:
\[
\text{Confidence Interval} = \left( \bar{x} - E, \bar{x} + E \right)
\]
where $ \bar{x} $ is the sample mean, which is 85.00.

Substitute the values:
\[
\text{Confidence Interval} = \left( 85.00 - 14.71, 85.00 + 14.71 \right)
\]
\[
\text{Confidence Interval} = \left( 70.29, 99.71 \right)
\]

### Interpretation
We are 90% confident that the true mean repair cost for the population of mobile devices lies between $70.29 and $99.71.

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**Final Answer:**
The 90% confidence interval for the population mean $ \mu $ is $ (70.29, 99.71) $.

Would you like a deeper explanation or have any questions?

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### Related Questions:
1. What would the confidence interval be at a 95% confidence level?
2. How would increasing the sample size affect the confidence interval width?
3. Why is the $ t $-distribution used here instead of the normal distribution?
4. What is the impact of the sample standard deviation on the margin of error?
5. How does changing the confidence level affect the critical $ t $-value?

**Tip:** Increasing the sample size reduces the margin of error, making the confidence interval narrower and more precise.

In a random sample of four mobile​ devices, the mean repair cost was ​$85.00 and the standard deviation was ​$12.50. 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.

Learn how to construct a 90% confidence interval for the population mean repair cost of mobile devices, using a t-distribution and sample data. This includes calculating the margin of error and interpreting the result with a 90% confidence level. Suitable for high school students in Grades 10-12.

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