SOLUTION: A sociologist found that in a sample of 50 retired men, the average number of jobs they had during their lifetimes was 7.2. The population standard deviation is 2.1. a) Find the b

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Question 1193183: A sociologist found that in a sample of 50 retired men, the average number of jobs they had during their lifetimes was 7.2. The population standard deviation is 2.1.
a) Find the best point estimate of the mean.
b) Find the 95% confidence interval of the mean number of jobs. c) Find the 99% confidence interval of the mean number of jobs.
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Answer by CPhill(1987)   (Show Source): You can put this solution on YOUR website!
**a) Best Point Estimate of the Mean**
* The best point estimate of the population mean is the sample mean.
* **Best Point Estimate = 7.2 jobs**
**b) 95% Confidence Interval**
1. **Find the critical value (z-score) for a 95% confidence level:**
* A 95% confidence level means there is 2.5% in each tail of the distribution.
* Using a standard normal distribution table or a calculator, the z-score for 97.5% (the area to the left of the right critical value) is approximately 1.96.
2. **Calculate the standard error of the mean:**
* Standard Error (SE) = σ / √n
* Where:
* σ is the population standard deviation (2.1)
* n is the sample size (50)
* SE = 2.1 / √50
* SE ≈ 0.297
3. **Calculate the margin of error:**
* Margin of Error = z-score * Standard Error
* Margin of Error = 1.96 * 0.297
* Margin of Error ≈ 0.58
4. **Calculate the confidence interval:**
* Lower Bound = Sample Mean - Margin of Error = 7.2 - 0.58 = 6.62
* Upper Bound = Sample Mean + Margin of Error = 7.2 + 0.58 = 7.78
* **95% Confidence Interval: (6.62 jobs, 7.78 jobs)**
**c) 99% Confidence Interval**
1. **Find the critical value (z-score) for a 99% confidence level:**
* A 99% confidence level means there is 0.5% in each tail of the distribution.
* Using a standard normal distribution table or a calculator, the z-score for 99.5% (the area to the left of the right critical value) is approximately 2.576.
2. **Calculate the margin of error:**
* Margin of Error = z-score * Standard Error
* Margin of Error = 2.576 * 0.297
* Margin of Error ≈ 0.766
3. **Calculate the confidence interval:**
* Lower Bound = Sample Mean - Margin of Error = 7.2 - 0.766 = 6.434
* Upper Bound = Sample Mean + Margin of Error = 7.2 + 0.766 = 7.966
* **99% Confidence Interval: (6.43 jobs, 7.97 jobs)**
**In summary:**
* **Best Point Estimate of the Mean:** 7.2 jobs
* **95% Confidence Interval:** (6.62 jobs, 7.78 jobs)
* **99% Confidence Interval:** (6.43 jobs, 7.97 jobs)

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