Each of 16 students measured the circumference of a tennis ball by four different methods, which were:
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.
The correct answer and explanation is :
a. Compute the mean measurement for each method:
The mean is calculated by summing the measurements and dividing by the total number of measurements.
For each method:
- Method A:
Sum = 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 = 380.9
Mean = 380.9 ÷ 16 = 23.81 cm - Method B:
Sum = 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 = 340.2
Mean = 340.2 ÷ 16 = 21.26 cm - Method C:
Sum = 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 = 335.7
Mean = 335.7 ÷ 16 = 20.98 cm - Method D:
Sum = 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 = 332.9
Mean = 332.9 ÷ 16 = 20.81 cm
b. Compute the median measurement for each method:
The median is the middle value when the measurements are sorted in increasing order. If there is an even number of measurements, the median is the average of the two middle values.
- Method A:
The middle values are 22.5 and 23.0.
Median = (22.5 + 23.0) ÷ 2 = 22.75 cm - Method B:
The middle values are 20.4 and 20.4.
Median = (20.4 + 20.4) ÷ 2 = 20.4 cm - Method C:
The middle values are 21.0 and 21.0.
Median = (21.0 + 21.0) ÷ 2 = 21.0 cm - Method D:
The middle values are 20.5 and 20.7.
Median = (20.5 + 20.7) ÷ 2 = 20.6 cm
c. Compute the 20% trimmed mean measurement for each method:
The 20% trimmed mean involves removing the lowest 20% and highest 20% of the data, then calculating the mean of the remaining data.
For 16 measurements, 20% of 16 is 3.2, so we remove the lowest 3 and the highest 3 measurements for each method.
- Method A:
Trimmed data: 20.0, 22.0, 22.0, 22.5, 23.0, 24.0, 24.0, 25.0, 25.0, 25.0, 25.0, 26.0
Trimmed sum = 286.5
Trimmed mean = 286.5 ÷ 12 = 23.875 cm - Method B:
Trimmed data: 19.6, 20.1, 20.4, 20.4, 20.4, 20.4, 20.5, 21.2, 22.0, 22.0, 22.0
Trimmed sum = 227.6
Trimmed mean = 227.6 ÷ 11 = 20.64 cm - Method C:
Trimmed data: 20.7, 20.8, 20.9, 21.0, 21.0, 21.0, 21.0, 21.0, 21.5, 21.5, 21.5
Trimmed sum = 229.9
Trimmed mean = 229.9 ÷ 11 = 20.99 cm - Method D:
Trimmed data: 20.0, 20.0, 20.2, 20.5, 20.5, 20.7, 20.7, 20.7, 21.0, 21.1, 21.5
Trimmed sum = 227.7
Trimmed mean = 227.7 ÷ 11 = 20.63 cm
d. Compute the first and third quartiles for each method:
The first quartile (Q1) is the median of the lower half, and the third quartile (Q3) is the median of the upper half.
- Method A:
Q1 = 20.5, Q3 = 25.0 - Method B:
Q1 = 19.6, Q3 = 22.0 - Method C:
Q1 = 20.8, Q3 = 21.5 - Method D:
Q1 = 20.2, Q3 = 21.2
e. Compute the standard deviation of the measurements for each method:
The formula for standard deviation is:
[ \text{SD} = \sqrt{\frac{1}{n} \sum_{i=1}^{n} (x_i – \mu)^2} ]
Where ( x_i ) are the individual measurements, ( \mu ) is the mean, and ( n ) is the number of measurements.
- Method A:
SD ≈ 2.47 cm - Method B:
SD ≈ 1.14 cm - Method C:
SD ≈ 0.30 cm - Method D:
SD ≈ 0.56 cm
f. For which method is the standard deviation the largest? Why should one expect this method to have the largest standard deviation?
The largest standard deviation is for Method A, with an SD of approximately 2.47 cm. This method involved estimating the circumference by eye, which can lead to greater variation in results due to subjective judgment and human error. Methods that rely on estimation or less precise tools tend to have larger standard deviations because the measurements are less consistent across participants.
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.
In general, it is better for a measurement method to have a smaller standard deviation. A smaller standard deviation indicates that the measurements are more consistent and reliable. When a measurement method produces results that are tightly clustered around the mean, it suggests that the method is precise and less prone to error. This is particularly important in scientific experiments, engineering, and other fields where accuracy and reproducibility are essential. A larger standard deviation, on the other hand, signifies more variability and less reliability, which could lead to incorrect conclusions or decisions. Therefore, consistency in measurement is usually preferred, and methods with smaller standard deviations are generally more trusted.