Answer :
To determine if the sample mean is an unbiased estimator, we need to verify that the mean of the sample means is equal to the population mean. Let's go through this step-by-step:
1. Identify the given ages: The ages of the five officers are 18, 18, 17, 16, and 15.
2. Calculate the population mean: The population mean (\(\mu\)) is given as 16.8.
3. List all possible samples of size 2 and their means:
- Sample: (18, 18), Sample Mean: 18
- Sample: (18, 17), Sample Mean: 17.5
- Sample: (18, 17), Sample Mean: 17.5
- Sample: (18, 16), Sample Mean: 17
- Sample: (18, 16), Sample Mean: 17
- Sample: (18, 15), Sample Mean: 16.5
- Sample: (18, 15), Sample Mean: 16.5
- Sample: (17, 16), Sample Mean: 16.5
- Sample: (17, 15), Sample Mean: 16
- Sample: (16, 15), Sample Mean: 15
4. Calculate the mean of the sample means:
[tex]\[ \text{Mean of the sample means} = \frac{18 + 17.5 + 17.5 + 17 + 17 + 16.5 + 16.5 + 16.5 + 16 + 15}{10} \][/tex]
Breaking this calculation down:
[tex]\[ 18 + 17.5 + 17.5 + 17 + 17 + 16.5 + 16.5 + 16.5 + 16 + 15 = 167.5 \][/tex]
Dividing this total by the number of samples (10):
[tex]\[ \text{Mean of the sample means} = \frac{167.5}{10} = 16.75 \][/tex]
5. Compare the mean of the sample means to the population mean:
The population mean is given as 16.8. The calculated mean of the sample means is 16.75.
Since the mean of the sample means is 16.75, which does not equal the population mean of 16.8, we conclude that the sample mean is not an unbiased estimator for the population mean.
Therefore, the correct answer is:
No, none of the sample means was 16.8; the sample mean is not an unbiased estimator.
1. Identify the given ages: The ages of the five officers are 18, 18, 17, 16, and 15.
2. Calculate the population mean: The population mean (\(\mu\)) is given as 16.8.
3. List all possible samples of size 2 and their means:
- Sample: (18, 18), Sample Mean: 18
- Sample: (18, 17), Sample Mean: 17.5
- Sample: (18, 17), Sample Mean: 17.5
- Sample: (18, 16), Sample Mean: 17
- Sample: (18, 16), Sample Mean: 17
- Sample: (18, 15), Sample Mean: 16.5
- Sample: (18, 15), Sample Mean: 16.5
- Sample: (17, 16), Sample Mean: 16.5
- Sample: (17, 15), Sample Mean: 16
- Sample: (16, 15), Sample Mean: 15
4. Calculate the mean of the sample means:
[tex]\[ \text{Mean of the sample means} = \frac{18 + 17.5 + 17.5 + 17 + 17 + 16.5 + 16.5 + 16.5 + 16 + 15}{10} \][/tex]
Breaking this calculation down:
[tex]\[ 18 + 17.5 + 17.5 + 17 + 17 + 16.5 + 16.5 + 16.5 + 16 + 15 = 167.5 \][/tex]
Dividing this total by the number of samples (10):
[tex]\[ \text{Mean of the sample means} = \frac{167.5}{10} = 16.75 \][/tex]
5. Compare the mean of the sample means to the population mean:
The population mean is given as 16.8. The calculated mean of the sample means is 16.75.
Since the mean of the sample means is 16.75, which does not equal the population mean of 16.8, we conclude that the sample mean is not an unbiased estimator for the population mean.
Therefore, the correct answer is:
No, none of the sample means was 16.8; the sample mean is not an unbiased estimator.