Question

1. consider the following information on ultimate tensile strength (lb/in) for a sample of ( n = 4 ) hard zirconium copper wire specimens (from “characterization methods for fine copper wire,” wire j. intl., aug., 1997: 74–80):

( \bar{x} = 76,831 ) ( s = 180 ) smallest ( x_i = 76,683 )
largest ( x_i = 77,048 )
determine the values of the two middle sample observations (and don’t do it by successive guessing!).

Explanation:

Step1: Recall the formula for the sample mean

The sample mean \(\bar{x}\) of a sample with \(n\) observations \(x_1, x_2, \dots, x_n\) is given by \(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}\). Here, \(n = 4\), so \(\sum_{i=1}^{4}x_i=n\bar{x}\).
We know that \(n = 4\) and \(\bar{x}=76831\), so \(\sum_{i = 1}^{4}x_i=4\times76831 = 307324\).

Step2: Let the four observations be ordered

Let the four observations be \(x_1 = 76683\) (smallest), \(x_2\), \(x_3\) (the two middle observations), and \(x_4=77048\) (largest). Then \(\sum_{i = 1}^{4}x_i=x_1 + x_2+x_3 + x_4\).

Step3: Substitute known values and solve for \(x_2 + x_3\)

Substitute \(x_1 = 76683\) and \(x_4 = 77048\) into the sum formula:
\(307324=76683+x_2 + x_3+77048\)
First, calculate \(76683 + 77048=153731\)
Then, \(x_2 + x_3=307324 - 153731=153593\)

Step4: Use the property of ordered data (median)

For a sample of size \(n = 4\) (even), the median is the average of the two middle observations, i.e., \(\text{Median}=\frac{x_2 + x_3}{2}\). Also, we know that the data is ordered, so \(x_1\leq x_2\leq x_3\leq x_4\).
We can also think about the fact that the sum of the two middle terms is \(153593\), and we can find the two middle terms by considering that the distribution of the data (since it's a sample of wire tensile strength, we can assume a somewhat symmetric or at least ordered such that we can find two numbers that add up to \(153593\) and are between \(76683\) and \(77048\)).
Let's denote \(x_2=a\) and \(x_3 = 153593 - a\). Since \(x_2\leq x_3\), we have \(a\leq153593 - a\), which implies \(2a\leq153593\) or \(a\leq76796.5\). Also, since \(x_2\geq x_1 = 76683\) and \(x_3\leq x_4 = 77048\), we have \(153593 - a\leq77048\), which implies \(a\geq153593 - 77048 = 76545\). But since \(x_2\geq x_1 = 76683\), we have \(76683\leq a\leq76796.5\) and \(x_3=153593 - a\) with \(76796.5\leq x_3\leq76910\) (wait, no, let's correct that: if \(a\leq76796.5\), then \(x_3 = 153593 - a\geq153593 - 76796.5 = 76796.5\), and since \(x_3\leq77048\), this is valid.
But we can also use the fact that for the sample, the two middle terms must satisfy \(x_2 + x_3=153593\) and we can find the two numbers by solving the system. Let's assume that the two middle terms are \(x_2\) and \(x_3\) such that \(x_2 + x_3=153593\) and we know that the data is ordered. Let's also note that the range of the data is \(77048 - 76683 = 365\), and the two middle terms are within this range around the mean.
The mean of the four data points is \(76831\), the mean of the two middle terms is \(\frac{153593}{2}=76796.5\).
Now, let's find the difference between the mean of the two middle terms and the smallest term: \(76796.5 - 76683 = 113.5\)
The difference between the largest term and the mean of the two middle terms: \(77048 - 76796.5 = 251.5\)
But maybe a better way is to set up the equation: let \(x_2=m\) and \(x_3 = 153593 - m\). Since the data is ordered, \(m\geq76683\) and \(153593 - m\leq77048\), so \(m\geq153593 - 77048 = 76545\) (but since \(m\geq76683\), we use \(m\geq76683\)). Also, \(m\leq153593 - m\) (since \(x_2\leq x_3\)) so \(m\leq76796.5\)
Now, let's consider that the sum of \(x_2\) and \(x_3\) is \(153593\). Let's try to find two numbers that add up to \(153593\) and are between \(76683\) and \(77048\).
Let's suppose that the two middle terms are \(x_2 = 76785\) and \(x_3 = 153593 - 76785 = 76808\)? Wait, no, let's do the calculation again.
Wait, \(x_1 = 76683\), \(x_4 = 77048\), sum of all four is \(4\times76831 = 307324\)
Sum of \(x_1\) and \(x_4\) is \(76683 + 77048 = 153731\)
Sum of \(x_2\) and \(x_3\) is \(307324 - 153731 = 153593\)
Now, let's let \(x_2 = a\) and \(x_3 = 153593 - a\), with \(76683\leq a\leq153593 - a\leq77048\)
So \(153593 - a\leq77048\implies a\geq153593 - 77048 = 76545\), but since \(a\geq76683\), we have \(76683\leq a\leq76796.5\) (because \(a\leq153593 - a\implies 2a\leq153593\implies a\leq76796.5\))
Now, let's check the difference between the mean of the two middle terms (\(76796.5\)) and the smallest term: \(76796.5 - 76683 = 113.5\)
The difference between the largest term and the mean of the two middle terms: \(77048 - 76796.5 = 251.5\)
But maybe we can find the two middle terms by considering that the sum is \(153593\). Let's try \(a = 76785\), then \(153593 - 76785 = 76808\). Let's check if these are between the smallest and largest: \(76683\leq76785\leq76808\leq77048\), yes.
Wait, but let's verify the sum: \(76683 + 76785 + 76808 + 77048\)
First, \(76683 + 76785 = 153468\), \(76808 + 77048 = 153856\), then \(153468+153856 = 307324\), which matches the total sum. Wait, no, that's not correct. Wait, \(76683+76785 = 153468\), \(76808 + 77048 = 153856\), and \(153468+153856 = 307324\), which is correct. But wait, the two middle terms are \(76785\) and \(76808\)? Wait, no, maybe I made a mistake in the assumption.
Wait, let's do it properly. Let the four numbers be \(a,b,c,d\) with \(a = 76683\), \(d = 77048\), \(n = 4\), \(\bar{x}=76831\), so \(a + b + c + d=4\times76831 = 307324\)
So \(b + c=307324 - a - d=307324 - 76683 - 77048\)
Calculate \(76683+77048 = 153731\)
Then \(b + c=307324 - 153731 = 153593\)
Now, since the data is ordered, \(a\leq b\leq c\leq d\), so \(b\geq a = 76683\) and \(c\leq d = 77048\)
Also, \(b\leq c\), so \(b\leq\frac{153593}{2}=76796.5\) and \(c\geq76796.5\)
Now, let's find two numbers \(b\) and \(c\) such that \(b + c = 153593\), \(76683\leq b\leq76796.5\), and \(76796.5\leq c\leq77048\)
Let's try \(b = 76785\), then \(c = 153593 - 76785 = 76808\). Check if \(76808\leq77048\): yes. Check if \(76785\geq76683\): yes. Check if \(76785\leq76808\): yes.
Wait, but let's check the sum again: \(76683+76785+76808+77048\)
\(76683+76785 = 153468\)
\(76808+77048 = 153856\)
\(153468+153856 = 307324\), which is correct.
But wait, is there a unique solution? Wait, no, actually, we can find the two middle terms by using the fact that the median is the average of the two middle terms, so median \(=\frac{b + c}{2}=\frac{153593}{2}=76796.5\)
So the two middle terms are equidistant from \(76796.5\) if the data is symmetric, but since it's a sample, they might not be, but in this case, since we have an integer sum (153593) and we need two integers (since tensile strength is likely an integer), we can find that \(b = 76796\) and \(c = 153593 - 76796 = 76797\)? Wait, no, \(76796+76797 = 153593\). Let's check:
\(76683+76796+76797+77048\)
\(76683+76796 = 153479\)
\(76797+77048 = 153845\)
\(153479+153845 = 307324\), which is correct.
Wait, but why the difference? Because I made a mistake earlier in assuming non - integer values, but tensile strength is likely measured as an integer, so the two middle terms should be integers that add up to 153593.
Let's check \(153593\) is odd, so the two middle terms must be one even and one odd or both non - integers, but since the given values (smallest and largest) are integers, it's possible that the middle terms are integers. Wait, \(76796+76797 = 153593\), and both are between \(76683\) and \(77048\), and \(76796\leq76797\), and \(76683\leq76796\) and \(76797\leq77048\).
Let's verify the mean: \(\frac{76683 + 76796+76797+77048}{4}=\frac{(76683+77048)+(76796+76797)}{4}=\frac{153731+153593}{4}=\frac{307324}{4}=76831\), which matches the given mean.
So the two middle observations are \(76796\) and \(76797\). Wait, but let's check the range: \(77048 - 76683 = 365\), and the two middle terms are \(76796\) and \(76797\), which are close to the mean of \(76831\).
Wait, another way: let the four numbers be \(x_1,x_2,x_3,x_4\) in order. We know that \(\sum x_i = 4\bar{x}=4\times76831 = 307324\)
We know \(x_1 = 76683\) and \(x_4 = 77048\), so \(x_2 + x_3=307324-(76683 + 77048)=307324 - 153731 = 153593\)
Let \(x_2 = m\), then \(x_3=153593 - m\)
Since \(x_1\leq x_2\leq x_3\leq x_4\), we have:
\(76683\leq m\leq153593 - m\leq77048\)
From \(153593 - m\leq77048\), we get \(m\geq153593 - 77048 = 76545\) (but since \(m\geq76683\), we use \(m\geq76683\))
From \(m\leq153593 - m\), we get \(m\leq76796.5\)
So \(m\) is in \([76683,76796.5]\) and \(x_3 = 153593 - m\) is in \([76796.5,77048]\)
Since the data is likely to be measured as integers, we can take \(m = 76796\) and \(x_3 = 153593 - 76796 = 76797\) (since \(76796+76797 = 153593\))
Let's check if these values are between the smallest and largest: \(76683\leq76796\leq76797\leq77048\), which is true.

Answer:

The two middle sample observations are \(\boldsymbol{76796}\) and \(\boldsymbol{76797}\)