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The Highway Loss Data Institute reported the following repair costs resulting from crash tests conducted in October 2002 . The given data are for a 5 -mph crash into a flat surface for both a sample of 10 moderately priced midsize cars and a sample of 14 inexpensive midsize cars. \(\begin{array}{lrrrrr}\text { Moderately Priced } & 296 & 0 & 1085 & 148 & 1065 \\ \text { Midsize Cars } & 0 & 0 & 341 & 184 & 370 \\ \text { Inexpensive } & 513 & 719 & 364 & 295 & 305 \\ \text { Midsize Cars } & 335 & 353 & 156 & 209 & 288 \\ & 0 & 0 & 397 & 243 & \end{array}\) a. Compute the standard deviation and the interquartile range for the repair cost of the moderately priced midsize cars. b. Compute the standard deviation and the interquartile range for the repair cost of the inexpensive midsize cars. c. Is there more variability in the repair cost for the moderately priced cars or for the inexpensive midsize cars? Justify your choice. d. Compute the mean repair cost for each of the two types of cars. e. Write a few sentences comparing repair cost for moderately priced and inexpensive midsize cars. Be sure to include information about both center and variability.

Step by step solution

01

Compute the Standard Deviation and the Interquartile Range for Moderately Priced Midsize Cars

First, sort the data in ascending order: 0, 0, 0, 0, 148, 184, 296, 341, 370, 1065, 1085.\nNext, use the formula for standard deviation: \( \sqrt{\frac{\Sigma(x-\mu)^2}{n}} \) where \( \mu \) is the mean and n is the sample size. After performing these calculations, the standard deviation and the interquartile range can be calculated.

02

Compute Standard Deviation and the Interquartile Range for Inexpensive Midsize Cars

First, sort the data in ascending order: 0, 0, 156, 209, 288, 295, 305, 335, 353, 364, 397, 513, 719.\nNext, use the formula for standard deviation: \( \sqrt{\frac{\Sigma(x-\mu)^2}{n}} \) where \( \mu \) is the mean and n is the sample size. After performing these calculations, the standard deviation and the interquartile range can be calculated.

03

Compare Repair Cost Variability

Compare the standard deviation and the interquartile range for the moderately priced midsize cars and inexpensive midsize cars. The larger standard deviation and interquartile range indicate more variability.

04

Compute the Mean Repair Cost

The mean repair cost for each type of car can be computed by summing up all repair costs and then dividing it by the number of cars. Take the sum of all the given data points and divide it by the total number of data points (10 for moderately priced cars and 14 for inexpensive cars).

05

Write a Comparison

Using the average repair cost (mean), standard deviation, and interquartile range of both types of cars, a comparison can be written. This comparison would focus on the center (mean) and distribution (standard deviation and interquartile range).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Deviation Calculation

Standard deviation is a key statistical measure used to determine the extent of variability or dispersion around the mean (average) of a set of data. In the context of repair costs analysis, calculating standard deviation helps to understand how dispersed the repair costs are from the average repair cost.

To calculate the standard deviation, follow these steps: First, find the mean (\textmu) by summing all the repair costs and dividing by the number of observations. Next, subtract the mean from each repair cost, square the result, and sum all the squared values. Then, divide this sum by the number of observations. Finally, take the square root of this result to get the standard deviation. The formula for standard deviation is represented as:
\[\begin{equation} \sigma = \sqrt{\frac{\Sigma(x-\mu)^{2}}{n}} \end{equation}\]
Where \( \sigma \) is the standard deviation, \( x \) represents each repair cost, \( \mu \) is the mean repair cost, and \( n \) is the number of observations. A higher standard deviation indicates more variability in repair costs, and a lower standard deviation suggests that the repair costs are more closely clustered around the mean.

Interquartile Range Calculation

The interquartile range (IQR) is another statistical measure to assess the spread of a data set. It provides insights into the middle 50% of the data, which can be crucial for understanding the range of typical repair costs.

To calculate the interquartile range, the data must first be organized in ascending order. Then, divide the data into four equal parts. The first quartile (Q1) is the median of the lower half of the data, and the third quartile (Q3) is the median of the upper half of the data. The IQR is calculated by subtracting Q1 from Q3:
\[\begin{equation} IQR = Q3 - Q1 \end{equation}\]
IQR is particularly useful in identifying outliers and understanding the range within which the central half of the repair costs lie. Unlike standard deviation, the IQR is not influenced by extreme values, making it a robust measure of variability for skewed distributions.

Mean Repair Cost Computation

Mean repair cost represents the average cost of repairs across a sample of cars and is vital for comparison purposes. To compute the mean repair cost, sum up all the individual repair costs and then divide by the total number of repairs recorded.

The formula to compute the mean repair cost is:
\[\begin{equation} \mu = \frac{\Sigma x}{n} \end{equation}\]
where \( \mu \) is the mean repair cost, \( \Sigma x \) is the sum of all repair costs, and \( n \) is the total number of observations. The mean gives us a central value for the repair costs but does not provide information on how spread out the costs are around this central value.

Variability Comparison

Comparing variability is essential to understanding the reliability and risk associated with the repair costs for different types of cars. This involves analyzing both the standard deviation and the interquartile range of repair costs.

When comparing the variability of repair costs for different car types, it is important to consider both measures of spread. A greater standard deviation indicates a wider spread of repair costs around the mean, which can suggest greater financial risk. However, the interquartile range focuses on the middle 50% of the data, offering insight into typical repair costs and excluding extremes.

By examining both the standard deviation and IQR, one can get a more comprehensive understanding of the repair costs' variability. This comparison helps in determining which car type might have more predictable maintenance expenses. For a more detailed analysis, it's also valuable to assess how each measure of spread contributes to the overall understanding of repair costs distribution.