Random samples of size 81 are taken from an infinite population whose mean and standard deviation are 45 and 9, respectively. The mean and standard error of the sampling distribution of the sample mean are:
Question
Random samples of size 81 are taken from an infinite population whose mean and standard deviation are 45 and 9, respectively. The mean and standard error of the sampling distribution of the sample mean are:
Solution 1
The mean and standard error of the sampling distribution of the sample mean can be calculated using the following formulas:
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Mean of the Sampling Distribution (μx̄): This is equal to the mean of the population (μ). So, μx̄ = μ = 45.
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Standard Error of the Sampling Distribution (σx̄): This is calculated by dividing the standard deviation of the population (σ) by the square root of the sample size (n). So, σx̄ = σ/√n = 9/√81 = 9/9 = 1.
So, the mean of the sampling distribution of the sample mean is 45 and the standard error of the sampling distribution of the sample mean is 1.
Solution 2
The mean and standard error of the sampling distribution of the sample mean can be calculated using the following steps:
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The mean of the sampling distribution of the sample mean (often denoted as μx̄) is equal to the mean of the population. So, μx̄ = μ = 45.
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The standard error of the sampling distribution of the sample mean (often denoted as σx̄) is calculated by dividing the standard deviation of the population (σ) by the square root of the sample size (n). So, σx̄ = σ/√n = 9/√81 = 9/9 = 1.
So, the mean of the sampling distribution of the sample mean is 45 and the standard error of the sampling distribution of the sample mean is 1.
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