Consider a population of 3 units having values 2, 4 and 6. A simple random sample (without replacement) of 2 units is to be drawn from the population. Let M denote the sample mean of this sample. Then which of the following statements are true?
Let the population be denoted by $P = \{2, 4, 6\}$. The population size is $N = 3$. A simple random sample (without replacement) of size $n = 2$ is drawn from this population. We are interested in the sample mean, denoted by $M$.
Possible Simple Random Samples
When drawing a simple random sample of size 2 without replacement from a population of 3 units, the possible samples are combinations of 2 units from the 3. The number of possible samples is given by $\binom{N}{n} = \binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2 \times 1}{(2 \times 1)(1)} = 3$.
The possible samples are:
- Sample 1: {2, 4}
- Sample 2: {2, 6}
- Sample 3: {4, 6}
Each sample has a probability of $\frac{1}{\binom{N}{n}} = \frac{1}{3}$ of being selected.
Calculating Sample Means
For each possible sample, we calculate the sample mean ($M$):
- For sample {2, 4}, the sample mean is $M = \frac{2+4}{2} = \frac{6}{2} = 3$.
- For sample {2, 6}, the sample mean is $M = \frac{2+6}{2} = \frac{8}{2} = 4$.
- For sample {4, 6}, the sample mean is $M = \frac{4+6}{2} = \frac{10}{2} = 5$.
The possible values of the sample mean $M$ are 3, 4, and 5, each with a probability of $\frac{1}{3}$.
Probability Distribution of M
We can list the possible values of $M$, $M^2$, $M^3$, and their corresponding probabilities:
| Sample | M | M2 | M3 | Probability |
|---|---|---|---|---|
| {2, 4} | 3 | \(3^2 = 9\) | \(3^3 = 27\) | \(1/3\) |
| {2, 6} | 4 | \(4^2 = 16\) | \(4^3 = 64\) | \(1/3\) |
| {4, 6} | 5 | \(5^2 = 25\) | \(5^3 = 125\) | \(1/3\) |
Evaluating Expected Value and Variance
Expected Value E(M)
The expected value of M, $E(M)$, is calculated as the sum of each possible value of M multiplied by its probability:
\[E(M) = \sum M \cdot P(M) = 3 \cdot \frac{1}{3} + 4 \cdot \frac{1}{3} + 5 \cdot \frac{1}{3}\] \[E(M) = \frac{3}{3} + \frac{4}{3} + \frac{5}{3} = \frac{3+4+5}{3} = \frac{12}{3} = 4\]Statement 1 says $E(M) = 4$. This statement is true.
Expected Value E(M2)
The expected value of M squared, $E(M^2)$, is calculated as the sum of each possible value of M squared multiplied by its probability:
\[E(M^2) = \sum M^2 \cdot P(M) = 9 \cdot \frac{1}{3} + 16 \cdot \frac{1}{3} + 25 \cdot \frac{1}{3}\] \[E(M^2) = \frac{9}{3} + \frac{16}{3} + \frac{25}{3} = \frac{9+16+25}{3} = \frac{50}{3}\]Statement 2 says $E(M^2) = 17$. Since $\frac{50}{3} \approx 16.67 \neq 17$, this statement is false.
Expected Value E(M3)
The expected value of M cubed, $E(M^3)$, is calculated as the sum of each possible value of M cubed multiplied by its probability:
\[E(M^3) = \sum M^3 \cdot P(M) = 27 \cdot \frac{1}{3} + 64 \cdot \frac{1}{3} + 125 \cdot \frac{1}{3}\] \[E(M^3) = \frac{27}{3} + \frac{64}{3} + \frac{125}{3} = \frac{27+64+125}{3} = \frac{216}{3} = 72\]Statement 3 says $E(M^3) = 72$. This statement is true.
Variance Var(M)
The variance of M, $Var(M)$, can be calculated using the formula $Var(M) = E(M^2) - [E(M)]^2$.
We have calculated $E(M) = 4$ and $E(M^2) = \frac{50}{3}$.
\[Var(M) = E(M^2) - [E(M)]^2 = \frac{50}{3} - (4)^2\] \[Var(M) = \frac{50}{3} - 16 = \frac{50}{3} - \frac{48}{3} = \frac{50-48}{3} = \frac{2}{3}\]Statement 4 says $Var(M) = 1$. Since $Var(M) = \frac{2}{3} \neq 1$, this statement is false.
Conclusion
Based on the calculations, the following statements are true:
- Statement 1: $E(M) = 4$ (True)
- Statement 3: $E(M^3) = 72$ (True)
Important Questions from Sampling
The process of converting the analog sample into discrete form is called ______.
Respondents of a recent sample survey provided names of friends they thought would be likely users of a new product. These friends were contacted, completed a survey, and asked to supply names of other likely users. Which method of sampling has been used in this survey ?
Which one among the following relates to the probability-based sampling technique?
In which process is the flat-top pulse amplitude modulated signal generated?
Let x(t) be a signal with Nyquist rate ω 0. Determine the Nyquist rate for y(t) = x(t)cos(ω 0t)
