A population distribution is very left skewed with mean μ = 40 and standard deviation σ = 10. Consider two sampling distributions from this population distribution. Sampling distribution #1 is created from the sample means from all possible random samples of size n = 9; sampling distribution #2 is created from the sample means from all possible random samples of size 81. How do the standard deviations compare? The standard deviation of sampling distribution #1 is _____________ the standard deviation of sampling distribution #2.
greater thanThe standard deviation of the sampling distribution of x¯ equals σ/n−√. The standard deviation for sampling distribution #1 is 10/9√ = 3.33 and the standard deviation for sampling distribution #2 is 10/81−−√ = 1.11.
The standard deviation of sampling distribution #1 will be larger than the standard deviation of sampling distribution #2.
We can use the formula for the standard deviation of a sampling distribution:
standard deviation of sampling distribution = standard deviation of population / square root of sample size
For sampling distribution #1, we have:
standard deviation of sampling distribution #1 = 10 / sqrt(9) = 10 / 3
For sampling distribution #2, we have:
standard deviation of sampling distribution #2 = 10 / sqrt(81) = 10 / 9
We can see that the standard deviation of sampling distribution #1 is larger since the denominator (sqrt(9)) is smaller than the denominator for sampling distribution #2 (sqrt(81)).
Therefore, the standard deviation of sampling distribution #1 is 3.33 (rounded to two decimal places) times larger than the standard deviation of sampling distribution #2.
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