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: Concept Overview:
The standard error of the estimate for the population mean represents the standard deviation of the sampling distribution of the sample mean, denoted as \(\bar{y}\). For Simple Random Sampling Without Replacement (SRSWOR) in a finite population, the calculation incorporates a finite population correction (FPC) factor.

Step 2: Core Formula:
The standard error (SE) of the sample mean \(\bar{y}\) is determined 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:- \(N\) represents the population size.- \(n\) represents the sample size.- \(S\) represents the population standard deviation.

Step 3: Step-by-Step Calculation:
Given:- Population size, \(N = 150\).- Sample size, \(n = 30\).- Population standard deviation, \(S = 11.9\).First, compute 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, determine the standard error by taking the square root of the variance:\[ \text{SE}(\bar{y}) = \sqrt{3.77626...} \approx 1.94326 \]This result is approximately 1.95.
Step 4: Final Result:
The standard error of the estimate for the population mean is approximately 1.95.