Chapter 3, Problem 7RE

Measure that ball: Each of 16 students measured the circumference of a tennis ball by two different methods:

A: Estimate the circumference by eye.

B: Measure the circumference by rolling the

The results (in centimeters) are given below, in increasing order for each method

1. Compute the sample standard deviation of the measurements for each method.
2. For which method is the sample standard deviation larger? Why should one expect this method to have the larger standard deviation?
3. 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?

To determine

To find: The sample standard deviation for each method.

Answer to Problem 7RE

Sample standard deviation for method A is2.87

Sample standard deviation for method B is 0.75

Explanation of Solution

For Method A:

Given data:

$x$
18
18
18
20
22
22
22.5
23
24
24
25
25
25
25
26
26.4

First, need to find the sample mean ( $\overline{x}$ ).

Formula:

  $\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{X}_i}}{n}$

Here n = 16 which is the count of measurements.

Values of $x_i$ are measurements.

Calculation:

  $\overline{x}=\frac{363.9}{16}=22.74$

Creating a table for finding population standard deviation:

$x$	$x-\overline{x}$	${(x-\overline{x})}^2$
18	-4.74	22.47
18	-4.74	22.47
18	-4.74	22.47
20	-2.74	7.51
22	-0.74	0.55
22	-0.74	0.55
22.5	-0.24	0.06
23	0.26	0.07
24	1.26	1.59
24	1.26	1.59
25	2.26	5.11
25	2.26	5.11
25	2.26	5.11
25	2.26	5.11
26	3.26	10.63
26.4	3.66	13.40
Sum		123.76

The formula of sample standard deviation:

  $s=\sqrt{\frac{{\displaystyle \sum {(x_i-\overline{x})}^2}}{n-1}}$

Calculation:

From the table,

  ${\displaystyle \sum {(x_i-\overline{x})}^2}=123.76$

Put in a formula,

  $s=\sqrt{\frac{123.76}{16}}=2.87$

Sample standard deviation = s = 2.87

For method B:

Given data

$x$
20
20
20
20
20.2
20.5
20.5
20.7
20.7
20.7
21
21.1
21.5
21.6
22.1
22.3

Calculated sample mean is,

  $\overline{x}=\frac{332.9}{16}=20.81$

Creating a table for finding population standard deviation:

$x$	$x-\overline{x}$	${(x-\overline{x})}^2$
20	-0.81	0.66
20	-0.81	0.66
20	-0.81	0.66
20	-0.81	0.66
20.2	-0.61	0.37
20.5	-0.31	0.10
20.5	-0.31	0.10
20.7	-0.11	0.01
20.7	-0.11	0.01
20.7	-0.11	0.01
21	0.19	0.04
21.1	0.29	0.08
21.5	0.69	0.48
21.6	0.79	0.62
22.1	1.29	1.66
22.3	1.49	2.22
Sum		8.33

Calculation:

From table,

  ${\displaystyle \sum {(x_i-\overline{x})}^2}=8.33$

Put in a formula,

  $s=\sqrt{\frac{8.33}{16}}=0.75$

Sample standard deviation = 0.75

To determine

To check: The standard deviation that is larger.

Answer to Problem 7RE

The standard deviation of method A is larger than method B.

Explanation of Solution

The standard deviation of method A is 2.87.

The standard deviation of method B is 0.75

From the above observation, it seems that the standard deviation of method A is larger than method B. As there is high variation in the measurement of method A which leads high standard deviation. The high standard deviation indicates that the data points are spread out over a large range of values.Method A is just by observational measurements which may lead to high variations in measurements. But method B is measuring the circumference by rolling the ball along the ruler which leads to lower the standard deviation

To determine

To justify the standard deviation that is better for the given methods.

Answer to Problem 7RE

Lower standard deviation is always better as the measurements tend to be very close to the mean.

Explanation of Solution

To expressing the variability of a population, the standard deviation is commonly used to measureconfidence in statistical conclusions.A low standard deviation indicates that the measurements tend to be very close to the mean and a high standard deviation indicates that the measurements are spread out over a large range of values.Method A is just by observational measurements which may lead to high variations in measurements. But method B is measuring the circumference by rolling the ball along the ruler which leads to lower the standard deviation.