Question 1170777: You are given the sample mean and the sample standard deviation Assume the population is normally distributed and use the T distribution define margine of error and construct a 95% confidence interval for the population mean. Interpret the results.
1. In a random sample of 8 people, the main commute time to work was 35.5 minutes and the standard deviation was 7.2 minutes.
2. In a random sample of 13 microwaves, the main repair cost was $80.00 and the standard deviation was $13.50.
3. In a random sample of 7 computers, the repair cost was $110.00 and a standard division was $44.50
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's solve each of these problems step by step.
**General Formula for Confidence Interval**
When the population standard deviation is unknown and the sample size is small (n < 30), we use the t-distribution.
Confidence Interval = Sample Mean ± Margin of Error
Margin of Error (E) = t_c * (s / √n)
Where:
* t_c = critical t-value (from t-distribution table)
* s = sample standard deviation
* n = sample size
**1. Commute Time**
* Sample Mean (x̄) = 35.5 minutes
* Sample Standard Deviation (s) = 7.2 minutes
* Sample Size (n) = 8
* Confidence Level = 95%
* **Degrees of Freedom (df):** df = n - 1 = 8 - 1 = 7
* **Critical t-value (t_c):** For a 95% confidence level and df = 7, t_c ≈ 2.365 (from a t-distribution table)
* **Margin of Error (E):** E = 2.365 * (7.2 / √8) ≈ 6.02 minutes
* **Confidence Interval:** 35.5 ± 6.02 = (29.48, 41.52) minutes
**Interpretation:** We are 95% confident that the true population mean commute time to work is between 29.48 and 41.52 minutes.
**2. Microwave Repair Cost**
* Sample Mean (x̄) = $80.00
* Sample Standard Deviation (s) = $13.50
* Sample Size (n) = 13
* Confidence Level = 95%
* **Degrees of Freedom (df):** df = n - 1 = 13 - 1 = 12
* **Critical t-value (t_c):** For a 95% confidence level and df = 12, t_c ≈ 2.179 (from a t-distribution table)
* **Margin of Error (E):** E = 2.179 * (13.50 / √13) ≈ $8.15
* **Confidence Interval:** $80.00 ± $8.15 = ($71.85, $88.15)
**Interpretation:** We are 95% confident that the true population mean repair cost for microwaves is between $71.85 and $88.15.
**3. Computer Repair Cost**
* Sample Mean (x̄) = $110.00
* Sample Standard Deviation (s) = $44.50
* Sample Size (n) = 7
* Confidence Level = 95%
* **Degrees of Freedom (df):** df = n - 1 = 7 - 1 = 6
* **Critical t-value (t_c):** For a 95% confidence level and df = 6, t_c ≈ 2.447 (from a t-distribution table)
* **Margin of Error (E):** E = 2.447 * (44.50 / √7) ≈ $41.06
* **Confidence Interval:** $110.00 ± $41.06 = ($68.94, $151.06)
**Interpretation:** We are 95% confident that the true population mean repair cost for computers is between $68.94 and $151.06.
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