Final answer:
The mean of the sampling distribution of the sample means will be 50 for both sampling with and without replacement. The standard error of the mean cannot be determined without the population variance, but it will be slightly smaller for sampling without replacement due to the finite population correction.
Step-by-step explanation:
The question asks to find the mean and standard error of the sampling distribution of means for a population with mean 50 and an unspecified variance, given a sample size of 20. There are two scenarios: sampling (i) without replacement and (ii) with replacement.
Firstly, since the population mean is given as 50, the mean of the sampling distribution of the sample means (also known as the expected value of the sample mean) will also be 50 for both scenarios because it is equal to the population mean.
The standard error of the mean (SEM), which is the standard deviation of the sample means, is normally calculated using the formula SEM = σ / √n, where σ is the population standard deviation and n is the sample size. Since the population variance is not specified, we cannot directly compute the SEM.
For sampling with replacement, the formula stays the same: SEM = σ / √n. However, for sampling without replacement, we must apply the finite population correction (FPC), which is √((N-n)/(N-1)), where N is the population size. Thus, the SEM without replacement is SEM = σ / √n * √((N-n)/(N-1)).
Without the population variance, the exact standard error cannot be computed for either scenario. Nevertheless, we can state that for sampling without replacement, the SEM will be slightly smaller due to the FPC factor, assuming the sample size is a small fraction of the population size. In both cases, the distribution of sample means for a sufficiently large sample size will approximate a normal distribution, centered around the population mean of 50.