Question 1179421: 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.
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 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 of 2: Find the critical value that should be used in constructing the confidence interval.**
* **Degrees of freedom (df):** n - 1 = 20 - 1 = 19
* **Confidence level:** 90% (0.90)
* Since the population standard deviation is unknown and the sample size is small (n < 30), we will use the t-distribution.
* We need to find the t-value that corresponds to a 90% confidence level with 19 degrees of freedom.
Using a t-table or calculator, we find the critical t-value to be approximately 1.729.
**Step 2 of 2: Construct the 90% confidence interval.**
* **Sample mean (x̄):** $57.22
* **Sample standard deviation (s):** $25.76
* **Sample size (n):** 20
* **Critical t-value (t*):** 1.729
1. **Calculate the standard error (SE):**
* SE = s / √n = 25.76 / √20 ≈ 5.76
2. **Calculate the margin of error (ME):**
* ME = t* * SE = 1.729 * 5.76 ≈ 9.96
3. **Construct the confidence interval:**
* Lower bound: x̄ - ME = 57.22 - 9.96 ≈ 47.26
* Upper bound: x̄ + ME = 57.22 + 9.96 ≈ 67.18
Therefore, the 90% confidence interval for the mean repair cost is approximately ($47.26, $67.18).
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