Which of the following is incorrect about the sampling distribution of the sample mean using the Central Limit Theorem?[Fill in the blank]

Question

BlackTom AI · Accepted Answer

Begin by identifying what the Central Limit Theorem (CLT) says about the sampling distribution of the sample mean. Under CLT, as n grows, the distribution of the sample mean tends toward a normal distribution with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by sqrt(n).
Option 1: 'As n increases, the sampling distribution of the sample mean approaches a normal distribution with the same mean and standard deviation as the population distribution.' This is not correct because while the distribution becomes approximately normal, the standard deviation is not the same as the population standard deviation; it shrinks to sigma/sqrt(n). Therefore, this option asserts equality of standard deviations incorrectly.
Option 2: 'The sampling distribution of the sample mean has a mean μ_xbar = μ' This is true: the mean of the sampling distribution equals the population mean.
Option 3: 'The standard deviation of the sampling distribution of the sample mean is smaller than the population standard deviation for a large n.' This is true because the standard deviation is sigma/sqrt(n), which is smaller than sigma for n > 1.
Option 4: 'The standard deviation of the sampling distribution of the sample mean is equal to the population standard deviation divided by the square root of n' This is exactly the CLT result, so it is true.
In summary, the incorrect statement is Option 1, since it claims the same standard deviation as the population rather than the reduced sigma/sqrt(n).

Which of the following is incorrect about the sampling distribution of the sample mean using the Central Limit Theorem?[Fill in the blank]

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