document.write( "Question 1194048: A population has a mean of µ = 100 and ơ = 15. Find the mean and standard deviation of the sampling distribution of samples means with sample size n.\r
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\n" ); document.write( "\n" ); document.write( "n = 15
\n" ); document.write( "n= 250
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Algebra.Com's Answer #826127 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "Regardless of the value of n, the mean of the sampling distribution is always gong to be mu = 100 in this case.\r
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\n" ); document.write( "\n" ); document.write( "If n = 15, then we have this standard deviation for the sampling distribution
\n" ); document.write( "sigma/sqrt(n)
\n" ); document.write( "15/sqrt(15)
\n" ); document.write( "3.87298
\n" ); document.write( "which is approximate.\r
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\n" ); document.write( "\n" ); document.write( "If n = 250, then,
\n" ); document.write( "sigma/sqrt(n)
\n" ); document.write( "15/sqrt(250)
\n" ); document.write( "0.94868
\n" ); document.write( "is the approximate standard deviation for the sampling distribution\r
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\n" ); document.write( "\n" ); document.write( "As n increases, the standard deviation for the sampling distribution decreases.
\n" ); document.write( "Intuitively it means that as the sample gets larger, we are narrowing in on the population parameter (there's less spread as the data is getting more consistent)\r
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\n" ); document.write( "\n" ); document.write( "Summary:
\n" ); document.write( "mean = 100 for each case
\n" ); document.write( "standard deviation = 3.87298 approximately when n = 15
\n" ); document.write( "standard deviation = 0.94868 approximately when n = 250
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