Descriptive Statistics Example - The American University ...



Descriptive Statistics Example

A manufacturer is investigating the operating life of laptop computer batteries. The following data are available.

|Life (min.) |Life (min.) |Life (min.) |Life (min.) |

|130 |145 |126 |146 |

|164 |130 |132 |152 |

|145 |129 |133 |155 |

|140 |127 |139 |137 |

|131 |126 |145 |148 |

|125 |132 |126 |126 |

|126 |135 |131 |129 |

|147 |136 |129 |136 |

|156 |146 |130 |146 |

|132 |142 |132 |132 |

Using the first two digits as stem we may develop the following plot:

| | | |Freq. |

| |12 |5 6 9 7 6 6 6 9 6 9 |10 |

| |13 |0 1 2 0 2 5 6 2 3 9 1 0 2 7 6 2 |16 |

| |14 |5 0 7 5 6 2 5 6 8 6 |10 |

| |15 |6 2 5 |3 |

| |16 |4 |1 |

Stem-and-leaf plot

The plot shows that most of the data is clustered around 130, with few data points crossing the 150 limit. One may conclude that the center of the data is somewhere in the 130s. Variation is harder to judge. Whether the variability is high or low can only be determined on a comparative basis at this stage. If another data set is available (may be for another brand), a back-to-back stem-and-leaf plot could be used to visually compare the variability in both sets.

By ordering the leafs, we get the following plot:

| | | |Freq |

| |12 |5 6 6 6 6 6 7 9 9 9 |10 |

| |13 |0 0 0 1 1 2 2 2 2 2 3 5 6 6 7 9 |16 |

| |14 |0 2 5 5 5 6 6 6 7 8 |10 |

| |15 |2 5 6 |3 |

| |16 |4 |1 |

Ordered Stem-and-leaf plot

From the plot above, we may determine many measures of dispersion and central tendency:

Minimum = 125, Maximum = 164, Range = 164 – 125 = 39.

Mode = 126, 132 (both are repeated 5 times- Bimodal data)

[pic]

Other measures require some calculations:

[pic]

These results confirm our initial conclusion that the center is in the 130s.

[pic]

Note: The average, median, mode, variance, and standard deviation may all be determined using the Excel functions AVERAGE, MEDIAN, MODE, VAR, and STDEV; respectively.

Also, we may use the ordered Stem-and-leaf plot (repeated below for convenience) to determine some probabilities:

| | | |Freq |

| |12 |5 6 6 6 6 6 7 9 9 9 |10 |

| |13 |0 0 0 1 1 2 2 2 2 2 3 5 6 6 7 9 |16 |

| |14 |0 2 5 5 5 6 6 6 7 8 |10 |

| |15 |2 5 6 |3 |

| |16 |4 |1 |

For example:

Only 3 observations are not less than 155. Therefore, P(X ................
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