Question

For the data sets in Exercises \(7-9,\) find the range, the sample variance and the sample standard deviation. The following data give the estimated prices of a 170 -gram can or a 200 -gram pouch of water-packed tuna for 14 different brands, based on prices paid nationally in supermarkets. $$ \begin{array}{rrrrrrr} .99 & 1.92 & 1.23 & .85 & .65 & .53 & 1.41 \\ 1.12 & .63 & .67 & .69 & .60 & .60 & .66 \end{array} $$

Short answer

Answer: The range is 1.39, the sample variance is 0.2437, and the sample standard deviation is 0.49.

Step-by-step solution

Arrange the data points in increasing order

List the prices in increasing order: $$ \begin{array}{rrrrrrr} 0.53 & 0.60 & 0.60 & 0.63 & 0.65 & 0.66 & 0.67 \\\ 0.69 & 0.85 & 0.99 & 1.12 & 1.23 & 1.41 & 1.92 \end{array} $$

Calculate the range

The range is the difference between the highest value and the lowest value in the data set. In this case, it is: Range = \(1.92 - 0.53 = 1.39\)

Find the mean price

To find the mean price, sum up all the prices and divide by the total number of data points which is 14: Mean Price (μ) = \(\frac{0.53 + 0.60 + ... + 1.41 + 1.92}{14} = \frac{12.76}{14} = 0.91\)

Compute the sample variance

The sample variance is given by the formula: $$s^2 = \frac{\sum_{i=1}^{n}(x_i - \mu)^2}{n-1}$$ where \(x_i\) represents each price, μ is the mean price, and n is the number of data points. In this case, n = 14. So, let's calculate the squared differences: $$ \begin{array}{cccccc} (0.53 - 0.91)^2 & (0.60 - 0.91)^2 & (0.60 - 0.91)^2 & (0.63 - 0.91)^2 & (0.65 - 0.91)^2 & ... \\\ (0.67 - 0.91)^2 & (0.69 - 0.91)^2 & (0.85 - 0.91)^2 & (0.99 - 0.91)^2 & (1.12 - 0.91)^2 & ... \\\ (1.23 - 0.91)^2 & (1.41 - 0.91)^2 & (1.92 - 0.91)^2 \end{array} $$ Now, calculate the sum of the squared differences: Sum of Squared Differences = \(0.1444 + 0.0961 + ... + 1.0201 = 3.1676\) So the sample variance (s^2) can be found by dividing the sum of squared differences by n-1 (13): Sample Variance (s^2) = \(\frac{3.1676}{13} = 0.2437\)

Calculate the sample standard deviation

The sample standard deviation (s) is the square root of the sample variance (s^2): Sample Standard Deviation (s) = \(\sqrt{0.2437} = 0.49\) So for this data set, the range is \(1.39\), the sample variance is \(0.2437\), and the sample standard deviation is \(0.49\).

Key concepts

Range

Range is the simplest measure of spread or dispersion in a data set. It helps us understand the spread between the smallest and the largest values in a dataset. To calculate the range, one only needs to subtract the smallest data value from the largest. This calculation gives an immediate sense of how spread out the data points are. The formula is simple:

• Range = Largest Value - Smallest Value

For the given data, the smallest value is 0.53 and the largest is 1.92. Thus, the range is calculated as follows:

• Range = 1.92 - 0.53 = 1.39.

Having a range of 1.39 suggests that there is notable variability in the prices of tuna cans or pouches across different brands. It is a helpful first indicator of how spread out the data might be.

Sample Variance

Sample variance measures how much the data points in a sample vary from the average or mean of the sample. Unlike the range, which only uses the extreme values, variance looks at all data points. Understanding the formula for sample variance is key to calculating it:

• Sample Variance (s²) = \( \frac{\sum_{i=1}^{n}(x_i - \mu)^2}{n-1} \)

• Where \( x_i \) are each of the data values, \( \mu \) is the mean of the sample, and \( n \) is the number of data points in the sample.

To find the sample variance, we first calculate the mean. Next, subtract the mean from each data value, square the result, and sum all these squared differences. Finally, divide this sum by one less than the number of data points the sample has. In this case, the mean of the sample is 0.91, and using the given data, the sum of squared differences came to 3.1676. Dividing this by 13 (since the sample size \( n \) is 14), the sample variance we get is approximately 0.2437. This tells us that there is some variation from the average price of tuna cans or pouches in these samples.

Sample Standard Deviation

Sample standard deviation builds upon variance and further helps in understanding the spread of data. It is the square root of the sample variance. While variance is measured in squared units, standard deviation provides a more usable measure since it has the same unit as the data. You can calculate it using the formula:

• Sample Standard Deviation (s) = \( \sqrt{s^2} \)

• Where \( s^2 \) is the sample variance.

For this data, we have already found a sample variance of 0.2437. Calculating its square root results in a sample standard deviation of approximately 0.49. The standard deviation offers valuable insight into the general data set by indicating how much each data point varies from the average, on a similar scale to the original data values. For this tuna price data, a standard deviation of 0.49 illustrates that most prices are relatively close to the average price, but there are still variations.