Problem 6 - Basiswissen Statistik: Kompaktkurs für Anwender aus Wirtschaft, Informatik und Technik

Question

Erstellen Sie ein Stamm-Blatt-Diagramm für die folgenden Messungen: $$ \begin{aligned} &11.3,9.82,9.81,9.2,6.87,7.4,7.56,7.67,8.23,8.43,8.55 \\ &9.12,10.2,10.43,9.99,11.12,10.82 \end{aligned} $$ Erstellen Sie auch eine geeignetes Histogramm. Geben Sie die zugehörige Häufigkeitsdichte an und berechnen Sie den zugehörigen Mittelwert und die Stichprobenvarianz.

Step-by-Step Solution

Answer

The stem-and-leaf plot is constructed, histogram is plotted, the mean is 9.29, and the sample variance is 2.34.

Step 1: Organize Data for Stem-and-Leaf Plot

First, sort the data in ascending order: \(6.87, 7.4, 7.56, 7.67, 8.23, 8.43, 8.55, 9.2, 9.12, 9.81, 9.82, 9.99, 10.2, 10.43, 10.82, 11.3, 11.12\). Identify the stem (integer part) and leaf (decimal part) for each measurement.

Step 2: Construct the Stem-and-Leaf Plot

Create the stem and leaf plot by listing the stem on the left and corresponding leaves on the right: - 6 | 8 - 7 | 4, 56, 67 - 8 | 23, 43, 55 - 9 | 12, 2, 81, 82, 99 - 10 | 2, 43, 82 - 11 | 3, 12.

Step 3: Classify Data for Histogram

Group the data into intervals: 6.5-7.5, 7.5-8.5, 8.5-9.5, 9.5-10.5, 10.5-11.5. Count how many data points fall into each interval: - 6.5-7.5: 2 - 7.5-8.5: 3 - 8.5-9.5: 1 - 9.5-10.5: 6 - 10.5-11.5: 3.

Step 4: Draw the Histogram

Plot the bars for each interval with the frequency on the y-axis. - 6.5-7.5 → Height 2 - 7.5-8.5 → Height 3 - 8.5-9.5 → Height 1 - 9.5-10.5 → Height 6 - 10.5-11.5 → Height 3.

Step 5: Calculate the Mean

Compute the mean by adding all data values and dividing by the count: Mean = \( \frac{6.87 + 7.4 + 7.56 + 7.67 + 8.23 + 8.43 + 8.55 + 9.12 + 9.2 + 9.81 + 9.82 + 9.99 + 10.2 + 10.43 + 10.82 + 11.3 + 11.12}{17} = 9.29 \).

Step 6: Compute Sample Variance

Calculate variance by finding the mean of squared differences from the mean. First, compute each \((x_i - \text{mean})^2\), sum them, and divide by one less than the count: Variance = \( \frac{\sum{(x_i - 9.29)^2}}{16} = 2.34 \).

Key Concepts

Histogram

A histogram is a graphical representation of data that uses bars to show the frequency of numerical data within certain intervals. In this exercise, data is divided into intervals such as 6.5-7.5, 7.5-8.5, and so on. For each interval, we determine how many data points fall within that range and plot a bar whose height represents this frequency.
For example, in this exercise, two values fall between 6.5 and 7.5, so the bar corresponding to this interval on the histogram has a height of 2. This process is repeated for other intervals:

• Three items fall between 7.5 and 8.5, so the bar height is 3.
• Only one item falls between 8.5 and 9.5, with a bar height of 1.
• Six items fall within 9.5 and 10.5, giving a bar height of 6, and so on.

Histograms provide a visual overview of the data distribution, allowing patterns like skewness and modality to be quickly identified.
Unlike bar graphs, histograms use continuous data, meaning the bars touch, representing the continuous nature of the data ranges.

Mean

The mean, also known as the arithmetic average, is a measure of central tendency that provides a central value for a dataset. In this context, you determine the mean by summing up all the measurements and dividing by the number of data points.
Given the data: \[ 6.87, 7.4, 7.56, 7.67, 8.23, 8.43, 8.55, 9.12, 9.2, 9.81, 9.82, 9.99, 10.2, 10.43, 10.82, 11.3, 11.12 \]Calculate the mean by adding these values and then dividing by 17, the number of measurements:

• Total sum = 158.03
• Mean = \( \frac{158.03}{17} = 9.29 \)

This result indicates that the average measurement is approximately 9.29. The mean summarizes the dataset with a single value, making it easy to compare with other data sets or expected outcomes.

Sample Variance

Sample variance is a measure that assesses how much the data points differ from the mean. It gives insight into the spread of the dataset, indicating whether the data points are close to or far from the mean.
To find it, first calculate the mean, which is 9.29 as determined earlier. Then, subtract the mean from each data point to find the difference. Square these differences to eliminate negative values and emphasize larger deviations.

• For instance, for a data point 6.87, the calculation is: \((6.87 - 9.29)^2 = 5.82 \)

Once all the squared differences are computed, sum them up. In this example, the sum of squared differences is 37.44.
Finally, divide this sum by \(n - 1\), where \(n\) is the number of data points (17 in this case), to get:\[ \text{Variance} = \frac{37.44}{16} = 2.34 \]This variance value of 2.34 reflects how spread out the values are around the mean.

Frequency Density

Frequency density is a statistical measure used in histograms to account for intervals of different widths. It is calculated by dividing the frequency of data points by the width of the interval, providing a way to fairly compare intervals of varying sizes.
In instances where all intervals have the same width, this calculation simplifies the appearance of the histogram. With equal interval widths as used in our histogram, the frequency itself can directly depict the data spread.

• For example, in the interval 6.5-7.5 containing 2 data points, with an interval width of 1 (7.5 - 6.5), the frequency density is \( \frac{2}{1} = 2 \).
• Similarly, for the interval 9.5-10.5 with 6 data points, the frequency density is \( \frac{6}{1} = 6 \).

This measure ensures that the heights of bars in the histogram indicate true density, critically useful if intervals have differing widths which was not the case here.