Question

If the standard deviation of marks obtained by 150 students is 11.9, then the standard error of the estimate of the population mean for a random sample of size 30 with SRSWOR, is:

• A. 1.87
• B. 1.95
• C. 1.78
• D. 2.15

Hint:

Don't forget the finite population correction (FPC) when sampling without replacement from a finite population, especially when the sample size \(n\) is a significant fraction of the population size \(N\) (a common rule of thumb is when \(n/N>0.05\)). Here, \(30/150 = 0.2\), so the FPC is essential.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The standard error of the estimate of the population mean is the standard deviation of the sampling distribution of the sample mean, \(\bar{y}\). For Simple Random Sampling Without Replacement (SRSWOR) from a finite population, the formula includes a finite population correction (FPC) factor.

Step 2: Key Formula or Approach:
The standard error (SE) of the sample mean \(\bar{y}\) is given by: \[ \text{SE}(\bar{y}) = \sqrt{\text{Var}(\bar{y})} = \sqrt{\frac{N-n}{N} \frac{S^2}{n}} = S \sqrt{\frac{N-n}{Nn}} \] Where: - \(N\) is the population size. - \(n\) is the sample size. - \(S\) is the population standard deviation.

Step 3: Detailed Explanation:
We are given the following values: - Population size, \(N = 150\). - Sample size, \(n = 30\). - Population standard deviation, \(S = 11.9\). First, calculate the finite population correction (FPC) factor: \[ \frac{N-n}{N} = \frac{150-30}{150} = \frac{120}{150} = 0.8 \] Next, calculate the variance of the sample mean: \[ \text{Var}(\bar{y}) = \frac{N-n}{N} \frac{S^2}{n} = 0.8 \times \frac{(11.9)^2}{30} = 0.8 \times \frac{141.61}{30} = 0.8 \times 4.72033... \approx 3.77626... \] Finally, calculate the standard error by taking the square root of the variance: \[ \text{SE}(\bar{y}) = \sqrt{3.77626...} \approx 1.94326 \] This value is closest to 1.95.
Step 4: Final Answer:
The standard error of the estimate of the population mean is approximately 1.95.