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The high costs in the California real estate market have caused families who cannot afford to buy bigger homes to consider backyard sheds as an alternative form of housing expansion. Many are using the backyard structures for home offices, art studios, and hobby areas as well as for additional storage. The mean price of a customized wooden, shingled backyard structure is \(\$ 3100\) ( Newsweek , September 29,2003 ). Assume that the standard deviation is \(\$ 1200\). a. What is the \(z\) -score for a backyard structure costing \(\$ 2300 ?\) b. What is the \(z\) -score for a backyard structure costing \(\$ 4900 ?\) c. Interpret the \(z\) -scores in parts (a) and (b). Comment on whether either should be considered an outlier. d. The Newsweek article described a backyard shed-office combination built in Albany, California, for \(\$ 13,000\). Should this structure be considered an outlier? Explain.

Step by step solution

01

Understanding the Data

We are given that the mean price for a backyard structure is \( \\(3100 \) and the standard deviation is \( \\)1200 \). These values will be used for calculating the \( z \)-scores.

02

Formula for Z-score

The formula to calculate the \( z \)-score is \( z = \frac{(X - \mu)}{\sigma} \), where \( X \) is the value of the observation, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

03

Calculating Z-score for $2300

Using the formula, we calculate the \( z \)-score for a structure costing \( \$2300 \) as follows: \[ z = \frac{2300 - 3100}{1200} = \frac{-800}{1200} = -0.67 \]

04

Calculating Z-score for $4900

Using the formula again for \( \$4900 \): \[ z = \frac{4900 - 3100}{1200} = \frac{1800}{1200} = 1.5 \]

05

Interpreting Z-scores for $2300 and $4900

A \( z \)-score indicates how many standard deviations an element is from the mean. The \( z \)-score of \(-0.67\) means \( \\(2300 \) is 0.67 standard deviations below the mean and \( z = 1.5 \) means \( \\)4900 \) is 1.5 standard deviations above the mean. Neither value is greater than \( 2 \) or less than \( -2 \), hence neither is an outlier.

06

Examining $13,000 for Outlier Status

Calculate the \( z \)-score for \( \$13,000 \) using the formula: \[ z = \frac{13000 - 3100}{1200} = \frac{9900}{1200} = 8.25 \]. A \( z \)-score of 8.25 indicates it is 8.25 standard deviations above the mean, which is typically considered an outlier since it is much greater than \( 3 \) or \( -3 \).

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Key Concepts

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

Understanding Standard Deviation

Standard deviation is a measure of how spread out numbers are in a dataset. It helps us understand how much variation exists from the average. In simpler terms, it tells us how much the prices of backyard sheds vary from the mean price of $3100. If the standard deviation is small, that means most shed prices are close to $3100. If it's large, the prices are more varied.

In our exercise, the standard deviation is $1200. This means that most backyard shed prices are expected to fall within $1200 above or below the average mean price. When you calculate a z-score, it's the number of these standard deviations a particular data point is from the mean. This is crucial for determining how common or rare a price point is in the dataset.

Calculating and Interpreting the Mean Price

The mean price is essentially the average price of all backyard sheds in the dataset. In this context, a mean price of $3100 is what you would expect to pay for a typical shed. When analyzing data, the mean is a central point which offers a quick snapshot of data tendencies.

The mean, however, can be affected by extreme values or outliers, so it’s always a good practice to also consider the standard deviation and look into the data distribution. In our example, when we calculated the z-scores for shed prices, we used this mean to determine how far off individual prices were from the average, which in turn helped us assess whether a price is typical or unusual.

Outlier Detection Using Z-Scores

Outlier detection is an important step in data analysis. It helps identify values that are significantly different from others in the dataset. In this exercise, we use z-scores to detect outliers.

• A z-score tells us how many standard deviations a value is from the mean.
• If a z-score is beyond \( \pm 2 \), it's often considered unusual.
• If a z-score goes beyond \( \pm 3 \), it is typically classified as an outlier.

In our case, the shed price of $13,000 had a z-score of 8.25, which means it's 8.25 standard deviations above the mean price. This indicates that this price is extremely rare within the dataset and should be considered an outlier. Outliers like these can skew the data analysis and might need special attention or exclusions depending on the objective of your analysis.