Download Degrees of Freedom: Ten values have a mean of 75

Degrees of Freedom: Ten values have a mean of 75.0. Nine of the values are
62,78,90,87,56,92,70,70, and 93.
a. Find the 10th value.
To find the mean (the average), you add up all the values, then divide the total by the
number of values that you have. We can use this to set up an equation and solve for the
missing value (we’ll call the missing value x).
75.0 = (62 + 78 + 90 + 87 + 56 + 92 + 70 + 70 + x)/10
750 = 62 + 78 + 90 + 87 + 56 + 92 + 70 + 70 + x
750 = 605 + x
145 = x
The missing value is 145. Now that we have all the values, find the mean of the 10
values to verify that it’s equal to 75.
b. We need to create a list of n values that have a specific known mean. We are free
to select any values we desire for some of the n values. How many of the n values
can be freely assigned before the remaining values are determined? (The result is
often referred to as the number of degrees of freedom.)
Whenever you have one group of n numbers, the degrees of freedom is equal to n – 1.
In the previous problem, we knew the values of n – 1 numbers (the 9 we were given) and
using this information, we were able to find the 10th number. What if we were given n –
2 numbers? Pretend that we have exactly the same information as before, but we don’t
know the value of the 9th number (let’s call it y). We can try to do the same calculations
to solve for the missing values:
75.0 = (62 + 78 + 90 + 87 + 56 + 92 + 70 + y + x)/10
750 = 62 + 78 + 90 + 87 + 56 + 92 + 70 + y + x
750 = 535 + y + x
215 = y + x
In this case, we can’t determine the values of the remaining numbers (for instance, we
could have y = 0 and x = 215, y = 200 and x = 15, or any of an infinite combination of
numbers). Because we can’t determine the remaining values when given n – 2 numbers,
the degrees of freedom has to be more than n – 2.
(Later on in statistics you might work with two groups of numbers, in which case you
would subtract 2 instead of 1.)
Range Rule of Thumb: Aluminum cans with a thickness of 0.0111 in. have axial
loads with a mean of 281.8 lb and a standard deviation of 27.8 lb. The axial load is
measured by applying pressure to the top of the can until it collapses. Use the range
rule of thumb to find the minimum and maximum "usual" axial loads. One
particular can had an axial load of 504 lb. Is that unusual?
A common rule of thumb is that the range of a data set is approximately 4 standard
deviations. Thus the maximum data element will be about 2 standard deviations above
the mean and the minimum data element about 2 standard deviations below the mean.
Using this rule, the maximum should be 2(27.8) = 55.6 lbs above the mean, and the
minimum should be 55.6 lbs below the mean:
maximum = 281.8 + 55.6 = 337.4 lbs
minimum = 282.8 – 55.6 = 226.2 lbs
If one can has an axial load of 504 lbs, this would be much higher than our estimated
maximum. Therefore, we can call it “unusual”.
(If you know more about the data set – i.e. if it is normally distributed – you can make
more exact estimates about whether a measurement is “usual” or not.)
Comparing Scores: Three students take equivalent stress tests. Which is the highest
relative score?
a. A score of 144 on a test with a mean of 128 and a standard deviation of 34.
b. A score of 90 on a test with a mean of 86 and a standard deviation of 18.
c. A score of 18 on a test with a mean of 15 and a standard deviation of 5.
You use standard deviations to measure how spread out a data set is. If the standard
deviation is relatively small, all the data are clumped tightly near the mean. If the
standard deviation is relatively large, the data are more spread out.
You can also use standard deviations to decribe individual data points. You can describe
each data point by saying how far it is from the mean in terms of standard deviations (i.e.
3 standard deviations above the mean, 1.4 standard deviations below the mean, etc. ). In
this problem, we’re going to convert each score into standard deviations. The score that
is the most standard deviations above the mean will be the highest relative score.
a. A score of 144 would be 16 points above the mean (144 – 128 = 16). If you divide 16
by 34, you can see that 16 is equal to 0.471 times 34 (16 = 0.471*34). This score is 0.471
standard deviations above the mean.
b. A score of 90 would be 4 points above the mean (90 – 86 = 4). If you divide 4 by 18,
you can see that 4 is equal to 0.222 times 18 (4 = 0.222*18). This score is 0.222 standard
deviations above the mean.
c. A score of 18 would be 3 points above the mean (18 – 15 = 3). If you divide 3 by 5,
you can see that 3 is equal to 0.6 times 18 (3 = 0.6*5). This score is 0.6 standard
deviations above the mean.
Therefore, the highest relative score (the one that is farthest above the mean in terms of
standard deviations) would be the one in part c.