SOLUTION: A consumer affairs investigator records the repair cost for 20 randomly selected refrigerators. A sample mean of $57.22 and standard deviation of $25.76 are subsequently computed.

Question 1179445: A consumer affairs investigator records the repair cost for 20 randomly selected refrigerators. A sample mean of $57.22 and standard deviation of $25.76 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the refrigerators. Assume the population is approximately normal.
Step 1 of 2: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.
Step 2 of 2: Construct the 90% confidence interval. Round your answer to two decimal places.
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
**Step 1: Find the critical value**
Since the population standard deviation is unknown, we use the t-distribution.
* Degrees of freedom (df) = n - 1 = 20 - 1 = 19
* Confidence level = 90%, so alpha (α) = 10% = 0.10
* We want the t-value that leaves 5% (α/2) in each tail.
Using a t-table or calculator with df = 19 and α/2 = 0.05, the critical value is approximately **1.729**.

**Step 2: Construct the confidence interval**
The formula for a confidence interval is:
```
Confidence Interval = Sample Mean ± (Critical Value) * (Standard Error)
```
Where the Standard Error (SE) = Sample Standard Deviation / √(Sample Size)
* Sample Mean = $57.22
* Critical Value = 1.729
* Sample Standard Deviation = $25.76
* Sample Size = 20
Let's calculate:
1. **Standard Error:** SE = $25.76 / √20 ≈ $5.77
2. **Margin of Error:** Margin of Error = (Critical Value) * (Standard Error) = 1.729 * $5.77 ≈ $9.97
3. **Confidence Interval:**
$57.22 ± $9.97
Lower Bound: $57.22 - $9.97 = $47.25
Upper Bound: $57.22 + $9.97 = $67.19
**Therefore, the 90% confidence interval for the mean repair cost is approximately ($47.25, $67.19).**
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