Each of 16 students measured the circumference of a tennis ball by four different methods, which were:

Question

Method A: Estimate the circumference by eye.
Method B: Measure the diameter with a ruler, and then compute the circumference.
Method C: Measure the circumference with a ruler and string.
Method D: Measure the circumference by rolling the ball along a ruler.
The results (in cm) are as follows, in increasing order for each method:
Method A: 18.0, 18.0, 18.0, 20.0, 22.0, 22.0, 22.5, 23.0, 24.0, 24.0, 25.0, 25.0, 25.0, 25.0, 26.0, 26.4.
Method B: 18.8, 18.9, 18.9, 19.6, 20.1, 20.4, 20.4, 20.4, 20.4, 20.5, 21.2, 22.0, 22.0, 22.0, 22.0, 23.6.
Method C: 20.2, 20.5, 20.5, 20.7, 20.8, 20.9, 21.0, 21.0, 21.0, 21.0, 21.0, 21.5, 21.5, 21.5, 21.5, 21.6.
Method D: 20.0, 20.0, 20.0, 20.0, 20.2, 20.5, 20.5, 20.7, 20.7, 20.7, 21.0, 21.1, 21.5, 21.6, 22.1, 22.3.
a. Compute the mean measurement for each method.
b. Compute the median measurement for each method.
c. Compute the 20% trimmed mean measurement for each method.
d. Compute the first and third quartiles for each method.
e. Compute the standard deviation of the measurements for each method.
f. For which method is the standard deviation the largest? Why should one expect
this method to have the largest standard deviation?
g. Other things being equal, is it better for a measurement method to have a smaller
standard deviation or a larger standard deviation? Or doesn’t it matter? Explain.

Answer 1

(a) The mean for A is

(18.0+18.0+18.0+20.0+22.0+22.0+22.5+23.0+24.0+24.0+25.0+25.0+25.0+25.0+26.0+26.4)?16 = 22.744

The mean for B is

(18.8+18.9+18.9+19.6+20.1+20.4+20.4+20.4+20.4+20.5+21.2+22.0+22.0+22.0+22.0+23.6)?16 = 20.700

The mean for C is

(20.2+20.5+20.5+20.7+20.8+20.9+21.0+21.0+21.0+21.0+21.0+21.5+21.5+21.5+21.5+21.6)?16 = 20.013

The mean for D is

(20.0+20.0+20.0+20.0+20.2+20.5+20.5+20.7+20.7+20.7+21.0+21.1+21.5+21.6+22.1+22.3)?16 = 20.806



(b) The median for A is (23.0 + 24.0)/2 = 23.5. The median for B is (20.4 + 20.4)/2 = 20.4. The median for C is(21.0 + 21.0)/2 = 21.0. The median for D is (20.7 + 20.7)/2

= 20.7.



(c) 0.20(16) = 3.2 ? 3. Trim the 3 highest and 3 lowest observations.

The 20% trimmed mean for A is

(20.0 + 22.0 + 22.0 + 22.5 + 23.0 + 24.0 + 24.0 + 25.0 + 25.0 + 25.0)?10 = 23.25

The 20% trimmed mean for B is

(19.6 + 20.1 + 20.4 + 20.4 + 20.4 + 20.4 + 20.5 + 21.2 + 22.0 + 22.0)?10 = 20.70

The 20% trimmed mean for C is

(20.7 + 20.8 + 20.9 + 21.0 + 21.0 + 21.0 + 21.0 + 21.0 + 21.5 + 21.5)?10 = 21.04

The 20% trimmed mean for D is

(20.0 + 20.2 + 20.5 + 20.5 + 20.7 + 20.7 + 20.7 + 21.0 + 21.1 + 21.5)?10 = 20.69



(d) 0.25(17) = 4.25. Therefore the first quartile is the average of the numbers in positions 4 and 5. 0.75(17) = 12.75. Therefore the third quartile is the average of the

numbers in positions 12 and 13.







(e) The variance for A is







The standard deviation for A is = 2.8724



The variance for B is





The standard deviation for B is = 1.3535.

The variance for C is





The standard deviation for C is 0.4193



The variance for D is





The standard deviation for D is = 0.7542



(f) Method A has the largest standard deviation. This could be expected, because of the four methods, this is the crudest. Therefore we could expect to see more variation in the way in which this method is carried out, resulting in more spread in the results.



(g) Other things being equal, a smaller standard deviation is better. With any measurement method, the result is somewhat different each time a measurement is made. When the standard deviation is small, a single measurement is more valuable, since we know that subsequent measurements would probably not be much different.