A population consists of the following four values: 12, 12, 14, and 16.
a. List all samples of size 2, and compute the mean of each sample.
b. Compute the mean of the distribution of the sample mean and the population mean. Compare the two values.
c. Compare the dispersion in the population with that of the sample mean.
The correct answer and explanation is :
Let’s go through each part step by step:
a. List all samples of size 2, and compute the mean of each sample.
The population consists of the values: 12, 12, 14, and 16. To find all possible samples of size 2, we list the combinations and calculate the mean of each sample.
The samples of size 2 are:
- (12, 12): Mean = (12 + 12) / 2 = 12
- (12, 14): Mean = (12 + 14) / 2 = 13
- (12, 16): Mean = (12 + 16) / 2 = 14
- (12, 14): Mean = (12 + 14) / 2 = 13
- (12, 16): Mean = (12 + 16) / 2 = 14
- (14, 16): Mean = (14 + 16) / 2 = 15
So, the sample means are: 12, 13, 14, 13, 14, and 15.
b. Compute the mean of the distribution of the sample mean and the population mean. Compare the two values.
- Population mean:
To compute the population mean, we sum all the population values and divide by the number of values:
[
\text{Population Mean} = \frac{12 + 12 + 14 + 16}{4} = \frac{54}{4} = 13.5
]
- Mean of the distribution of the sample mean:
The mean of the sample means is the average of the sample means we calculated earlier:
[
\text{Mean of Sample Means} = \frac{12 + 13 + 14 + 13 + 14 + 15}{6} = \frac{81}{6} = 13.5
]
As we can see, both the population mean and the mean of the sample means are equal to 13.5. This is an example of the sampling distribution of the sample mean being an unbiased estimator of the population mean.
c. Compare the dispersion in the population with that of the sample mean.
To compare the dispersion, we calculate the variance and standard deviation of both the population and the sample means.
- Population variance:
First, we calculate the variance of the population. The formula for variance is:
[
\text{Population Variance} = \frac{1}{N} \sum_{i=1}^{N} (x_i – \mu)^2
]
Where (x_i) is each data point and (\mu) is the population mean.
[
\text{Population Variance} = \frac{(12 – 13.5)^2 + (12 – 13.5)^2 + (14 – 13.5)^2 + (16 – 13.5)^2}{4}
]
[
= \frac{(1.5)^2 + (1.5)^2 + (0.5)^2 + (2.5)^2}{4}
]
[
= \frac{2.25 + 2.25 + 0.25 + 6.25}{4} = \frac{11}{4} = 2.75
]
Thus, the population variance is 2.75. The population standard deviation is:
[
\text{Population Standard Deviation} = \sqrt{2.75} \approx 1.66
]
- Variance of the distribution of the sample mean:
The variance of the sample mean is the population variance divided by the sample size (since we’re using a sample of size 2):
[
\text{Sample Mean Variance} = \frac{\text{Population Variance}}{n} = \frac{2.75}{2} = 1.375
]
So, the variance of the sample means is 1.375, and the standard deviation of the sample means is:
[
\text{Standard Deviation of Sample Means} = \sqrt{1.375} \approx 1.17
]
Comparison:
- The population variance is 2.75, while the variance of the sample means is 1.375.
- The population standard deviation is approximately 1.66, while the standard deviation of the sample means is approximately 1.17.
As expected, the dispersion (measured by variance and standard deviation) of the sample means is less than that of the population. This is because, when sampling, there is less variability in the sample means compared to individual values in the population. This phenomenon is known as the law of large numbers, which states that as the sample size increases, the sample mean will tend to get closer to the population mean.
In this case, using a sample size of 2 results in a reduction in dispersion, but the difference is not as dramatic as it would be with larger sample sizes.