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Find the following:
\nSample Mean = [[0]]
\nSample Standard Deviation = [[1]]
\nSample Standard Error = [[4]]
\nMedian = [[2]] Give the exact value
\nInterquartile Range = [[3]] Give the exact value
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\nLet the data be donated $x$, then the sample mean $\\bar{x}$ is
\n\\begin{align}
\\bar{x} &= \\frac{\\var{r0[0]}+\\var{r0[1]}+\\var{r0[2]}+\\var{r0[3]}+\\var{r0[4]}+\\var{r0[5]}}{\\var{length(r0)}} \\\\
&= \\frac{\\var{sum(r0)}}{\\var{length(r0)}} \\\\
&= \\var{mn}\\text{,}
\\end{align}
to 2 decimal places.
\n\nSample standard deviation and standard error:
\nThe sample standard deviation, $s$, is given by
\n\\begin{align}
s &= \\sqrt{\\frac{\\sum{(x-\\bar{x})^2}}{n-1}} \\\\
&= \\sqrt{\\frac{(\\var{r0[0]}-\\var{mn})^2+(\\var{r0[1]}-\\var{mn})^2+(\\var{r0[2]}-\\var{mn})^2+(\\var{r0[3]}-\\var{mn})^2+(\\var{r0[4]}-\\var{mn})^2+(\\var{r0[5]}-\\var{mn})^2}{\\var{length(r0)-1}}} \\\\
&= \\sqrt{\\frac{\\var{(r0[0]-mn)^2}+\\var{(r0[1]-mn)^2}+\\var{(r0[2]-mn)^2}+\\var{(r0[3]-mn)^2}+\\var{(r0[4]-mn)^2}+\\var{(r0[5]-mn)^2}}{\\var{length(r0)-1}}} \\\\
&= \\sqrt{\\frac{\\var{variance(r0,true)*mn}}{\\var{length(r0)-1}}} \\\\
&= \\sqrt{\\var{variance(r0,true)}} \\\\
&= \\var{dpformat(stdev(r0,true),2)}\\text{.}
\\end{align}
The sample standard error is the sample standard deviation divided by the square root of the sample size.
\n$\\text{Sample standard error} = \\displaystyle{\\frac{\\var{stdev}}{\\sqrt{\\var{n}}} =\\var{dpformat(stderr,2)}}$.
\nMedian and Interquartile range:
\nIf you order the data in increasing order you get the following table:
\n$\\var{r1[0]}$ | \n$\\var{r1[1]}$ | \n$\\var{r1[2]}$ | \n$\\var{r1[3]}$ | \n$\\var{r1[4]}$ | \n$\\var{r1[5]}$ | \n
Denote the ordered data by $x_j$, thus $x_{3}=\\var{r1[2]}$ for example.
\nThe median lies between the 2nd and 3rd entries in the ordered table and is given by:
\n\\[0.5\\times x_{3}+0.5\\times x_{4} = 0.5\\times\\var{r1[2]}+0.5\\times \\var{r1[3]}=\\var{median}\\]
\nThe lower quartile is given by $x_2 = \\var{r1[1]}$
\nThe upper quartile is given by $x_5 = \\var{r1[4]}$
\nThe interquartile range is defined to be
\n\\[ \\text{Upper Quartile} – \\text{Lower Quartile} \\]
\nand so in this case we have:
\n\\[ \\text{Interquartile range} = \\var{uquartile}-\\var{lquartile}=\\var{interq} \\]
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\n$\\var{r0[0]}$ | \n$\\var{r0[1]}$ | \n$\\var{r0[2]}$ | \n$\\var{r0[3]}$ | \n$\\var{r0[4]}$ | \n$\\var{r0[5]}$ | \n
Sample of size $24$ is given in a table. Find sample mean, sample standard deviation, sample median and the interquartile range.
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