https://file.aiquickdraw.com/mathbot/file/1729786440722z07cglfv.jpg

The problem asks for the construction of a 95% confidence interval for the average repair cost ($\mu$) of DVD players, using data from a random sample. 

The key information provided is:
- Sample size ($n$) = 16
- Sample mean ($\bar{x}$) = \$50
- Sample standard deviation ($s$) = \$10
- Confidence level = 95%

Since the sample size is small ($n < 30$) and the population standard deviation is unknown, we should use the *t-distribution* to calculate the confidence interval.

The general formula for the confidence interval using the t-distribution is:

\[
\left( \bar{x} - t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}, \bar{x} + t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} \right)
\]

Where:
- $\bar{x}$ is the sample mean,
- $t_{\alpha/2}$ is the critical value from the t-distribution,
- $s$ is the sample standard deviation,
- $n$ is the sample size.

Now, looking at the multiple-choice options in your image, the correct answer matches this format. The correct option is:

\[
\left( \bar{x} - t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}, \bar{x} + t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} \right)
\]

This is the **first option** provided in the image.

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Would you like further details on how to calculate the critical value or more practice problems?

Here are 5 questions that might help expand your understanding:
1. How is the t-distribution different from the z-distribution in terms of confidence intervals?
2. What happens to the confidence interval if the sample size increases?
3. Why do we use the sample standard deviation ($s$) instead of the population standard deviation ($\sigma$) here?
4. How does the confidence level affect the width of the confidence interval?
5. If you were constructing a 99% confidence interval, how would the critical value change?

**Tip**: Always ensure you know whether to use a t-distribution or z-distribution based on the sample size and whether the population standard deviation is known.

In a random sample of 16 DVD players brought in for repairs, the average repair cost was $50 and the standard deviation was $10. Assume that the repair costs are normally distributed. Construct a 95% confidence interval for μ, where μ represents the average repair cost for DVD players. Which equation should use to construct a confidence interval for average repair cost?

Learn how to construct a 95% confidence interval for the average repair cost of DVD players using a small sample size and the t-distribution. This problem involves key statistical concepts such as sample means, confidence levels, and the use of t-distribution for inference.

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