Statistics Examples

8

$8$ ,

9

$9$ ,

0

$0$ ,

7

$7$ ,

8

$8$ ,

9

$9$ ,

6

$6$ ,

7

$7$ ,

0

$0$ ,

3

$3$ ,

4

$4$

Step 1

The number of classes can be estimated using the rounded output of Sturges' rule,

N = 1 + 3.322 \log (n)

$N = 1 + 3.322 \log (n)$ , where

N

$N$ is the number of classes and

n

$n$ is the number of items in the data set.

1 + 3.322 \log (7) = 3.80741568

$1 + 3.322 \log (7) = 3.80741568$

Step 2

Select

4

$4$ classes for this example.

4

$4$

Step 3

Find the data range by subtracting the minimum data value from the maximum data value. In this case, the data range is

9 - 0 = 9

$9 - 0 = 9$ .

9

$9$

Step 4

Find the class width by dividing the data range by the desired number of groups. In this case,

\frac{9}{4} = 2.25

$\frac{9}{4} = 2.25$ .

2.25

$2.25$

Step 5

Round

2.25

$2.25$ up to the nearest whole number. This will be the size of each group.

3

$3$

Step 6

Start with

0

$0$ and create

4

$4$ groups of size

3

$3$ .

\begin{array}{c} Class & Class Boundaries & Frequency \\ 0 - 2 \\ 3 - 5 \\ 6 - 8 \\ 9 - 11 \end{array}

$\begin{array}{c} Class & Class Boundaries & Frequency \\ 0 - 2 \\ 3 - 5 \\ 6 - 8 \\ 9 - 11 \end{array}$

Step 7

Determine the class boundaries by subtracting

0.5

$0.5$ from the lower class limit and by adding

0.5

$0.5$ to the upper class limit.

\begin{array}{c} Class & Class Boundaries & Frequency \\ 0 - 2 & - 0.5 - 2.5 \\ 3 - 5 & 2.5 - 5.5 \\ 6 - 8 & 5.5 - 8.5 \\ 9 - 11 & 8.5 - 11.5 \end{array}

$\begin{array}{c} Class & Class Boundaries & Frequency \\ 0 - 2 & - 0.5 - 2.5 \\ 3 - 5 & 2.5 - 5.5 \\ 6 - 8 & 5.5 - 8.5 \\ 9 - 11 & 8.5 - 11.5 \end{array}$

Step 8

Draw a tally mark next to each class for each value that is contained within that class.

\begin{array}{c} Class & Class Boundaries & Frequency \\ 0 - 2 & - 0.5 - 2.5 & | | \\ 3 - 5 & 2.5 - 5.5 & | | \\ 6 - 8 & 5.5 - 8.5 & | | | | | \\ 9 - 11 & 8.5 - 11.5 & | | \end{array}

$\begin{array}{c} Class & Class Boundaries & Frequency \\ 0 - 2 & - 0.5 - 2.5 & | | \\ 3 - 5 & 2.5 - 5.5 & | | \\ 6 - 8 & 5.5 - 8.5 & | | | | | \\ 9 - 11 & 8.5 - 11.5 & | | \end{array}$

Step 9

Count the tally marks to determine the frequency of each class.

\begin{array}{c} Class & Class Boundaries & Frequency \\ 0 - 2 & - 0.5 - 2.5 & 2 \\ 3 - 5 & 2.5 - 5.5 & 2 \\ 6 - 8 & 5.5 - 8.5 & 5 \\ 9 - 11 & 8.5 - 11.5 & 2 \end{array}

$\begin{array}{c} Class & Class Boundaries & Frequency \\ 0 - 2 & - 0.5 - 2.5 & 2 \\ 3 - 5 & 2.5 - 5.5 & 2 \\ 6 - 8 & 5.5 - 8.5 & 5 \\ 9 - 11 & 8.5 - 11.5 & 2 \end{array}$

Step 10

The relative frequency of a data class is the percentage of data elements in that class. The relative frequency can be calculated using the formula

f_{i} = \frac{f}{n}

$f_{i} = \frac{f}{n}$ , where

f

$f$ is the absolute frequency and

n

$n$ is the sum of all frequencies.

f_{i} = \frac{f}{n}

$f_{i} = \frac{f}{n}$

Step 11

n

$n$ is the sum of all frequencies. In this case,

n = 2 + 2 + 5 + 2 = 11

$n = 2 + 2 + 5 + 2 = 11$ .

n = 11

$n = 11$

Step 12

The relative frequency can be calculated using the formula

f_{i} = \frac{f}{n}

$f_{i} = \frac{f}{n}$ .

\begin{array}{c} Class & Class Boundaries & Frequency (f) & f_{i} \\ 0 - 2 & - 0.5 - 2.5 & 2 & \frac{2}{11} \\ 3 - 5 & 2.5 - 5.5 & 2 & \frac{2}{11} \\ 6 - 8 & 5.5 - 8.5 & 5 & \frac{5}{11} \\ 9 - 11 & 8.5 - 11.5 & 2 & \frac{2}{11} \end{array}

$\begin{array}{c} Class & Class Boundaries & Frequency (f) & f_{i} \\ 0 - 2 & - 0.5 - 2.5 & 2 & \frac{2}{11} \\ 3 - 5 & 2.5 - 5.5 & 2 & \frac{2}{11} \\ 6 - 8 & 5.5 - 8.5 & 5 & \frac{5}{11} \\ 9 - 11 & 8.5 - 11.5 & 2 & \frac{2}{11} \end{array}$

Step 13

Simplify the relative frequency column.

\begin{array}{c} Class & Class Boundaries & Frequency (f) & f_{i} \\ 0 - 2 & - 0.5 - 2.5 & 2 & 0.\overline{18} \\ 3 - 5 & 2.5 - 5.5 & 2 & 0.\overline{18} \\ 6 - 8 & 5.5 - 8.5 & 5 & 0.\overline{45} \\ 9 - 11 & 8.5 - 11.5 & 2 & 0.\overline{18} \end{array}

$\begin{array}{c} Class & Class Boundaries & Frequency (f) & f_{i} \\ 0 - 2 & - 0.5 - 2.5 & 2 & 0.\overline{18} \\ 3 - 5 & 2.5 - 5.5 & 2 & 0.\overline{18} \\ 6 - 8 & 5.5 - 8.5 & 5 & 0.\overline{45} \\ 9 - 11 & 8.5 - 11.5 & 2 & 0.\overline{18} \end{array}$

Step 14

Multiply every relative frequency by

100

$100$ to get the percentage frequency.

\begin{array}{c} Class & Class Boundaries & Frequency (f) & f_{i} & Percent \\ 0 - 2 & - 0.5 - 2.5 & 2 & 0.\overline{18} & 0.\overline{18} \cdot 100 % \\ 3 - 5 & 2.5 - 5.5 & 2 & 0.\overline{18} & 0.\overline{18} \cdot 100 % \\ 6 - 8 & 5.5 - 8.5 & 5 & 0.\overline{45} & 0.\overline{45} \cdot 100 % \\ 9 - 11 & 8.5 - 11.5 & 2 & 0.\overline{18} & 0.\overline{18} \cdot 100 % \end{array}

$\begin{array}{c} Class & Class Boundaries & Frequency (f) & f_{i} & Percent \\ 0 - 2 & - 0.5 - 2.5 & 2 & 0.\overline{18} & 0.\overline{18} \cdot 100 % \\ 3 - 5 & 2.5 - 5.5 & 2 & 0.\overline{18} & 0.\overline{18} \cdot 100 % \\ 6 - 8 & 5.5 - 8.5 & 5 & 0.\overline{45} & 0.\overline{45} \cdot 100 % \\ 9 - 11 & 8.5 - 11.5 & 2 & 0.\overline{18} & 0.\overline{18} \cdot 100 % \end{array}$

Step 15

Simplify the Percentage column.

\begin{array}{c} Class & Class Boundaries & Frequency (f) & f_{i} & Percent \\ 0 - 2 & - 0.5 - 2.5 & 2 & 0.\overline{18} & 18.\overline{18} % \\ 3 - 5 & 2.5 - 5.5 & 2 & 0.\overline{18} & 18.\overline{18} % \\ 6 - 8 & 5.5 - 8.5 & 5 & 0.\overline{45} & 45.\overline{45} % \\ 9 - 11 & 8.5 - 11.5 & 2 & 0.\overline{18} & 18.\overline{18} % \end{array}

$\begin{array}{c} Class & Class Boundaries & Frequency (f) & f_{i} & Percent \\ 0 - 2 & - 0.5 - 2.5 & 2 & 0.\overline{18} & 18.\overline{18} % \\ 3 - 5 & 2.5 - 5.5 & 2 & 0.\overline{18} & 18.\overline{18} % \\ 6 - 8 & 5.5 - 8.5 & 5 & 0.\overline{45} & 45.\overline{45} % \\ 9 - 11 & 8.5 - 11.5 & 2 & 0.\overline{18} & 18.\overline{18} % \end{array}$

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