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How do grocery costs compare across the country? Using a market basket of 10 items including meat, milk, bread, eggs, coffee, potatoes, cereal, and orange juice, Where to Retire magazine calculated the cost of the market basket in six cities and in six retirement areas across the country (Where to Retire, November/December 2003 ). The data with market basket cost to the nearest dollar are as follows: $$\begin{array}{lclr} \text { City } & \text { cost } & \text { Retirement Area } & \text { cost } \\\ \text { Buffalo, NY } & \$ 33 & \text { Biloxi-Gulfport, MS } & \$ 29 \\ \text { Des Moines, IA } & 27 & \text { Asheville, NC } & 32 \\ \text { Hartford, CT } & 32 & \text { Flagstaff, AZ } & 32 \\ \text { Los Angeles, CA } & 38 & \text { Hilton Head, SC } & 34 \\ \text { Miami, FL } & 36 & \text { Fort Myers, FL } & 34 \\ \text { Pittsburgh, PA } & 32 & \text { Santa Fe, NM } & 31 \end{array}$$ a. Compute the mean, variance, and standard deviation for the sample of cities and the sample of retirement areas. b. What observations can be made based on the two samples?

Step by step solution

01

List the Costs

First, identify and list the costs for both cities and retirement areas from the data given. Cities: Buffalo, NY: $33, Des Moines, IA: $27, Hartford, CT: $32, Los Angeles, CA: $38, Miami, FL: $36, Pittsburgh, PA: $32. Retirement Areas: Biloxi-Gulfport, MS: $29, Asheville, NC: $32, Flagstaff, AZ: $32, Hilton Head, SC: $34, Fort Myers, FL: $34, Santa Fe, NM: $31.

02

Calculate the Mean for Cities

Sum the costs for all cities and divide by the number of cities to find the mean.Mean for cities: \[\text{Mean} = \frac{33 + 27 + 32 + 38 + 36 + 32}{6} = \frac{198}{6} = 33\]

03

Calculate the Mean for Retirement Areas

Sum the costs for all retirement areas and divide by the number of areas to find the mean.Mean for retirement areas:\[\text{Mean} = \frac{29 + 32 + 32 + 34 + 34 + 31}{6} = \frac{192}{6} = 32\]

04

Calculate the Variance for Cities

Compute the variance by finding the average of squared differences from the mean.For cities:\[\text{Variance} = \frac{(33 - 33)^2 + (27 - 33)^2 + (32 - 33)^2 + (38 - 33)^2 + (36 - 33)^2 + (32 - 33)^2}{6}\]\[= \frac{0 + 36 + 1 + 25 + 9 + 1}{6} = \frac{72}{6} = 12\]

05

Calculate the Variance for Retirement Areas

Compute the variance for the retirement areas using a similar method as for cities.Variance for retirement areas:\[\text{Variance} = \frac{(29 - 32)^2 + (32 - 32)^2 + (32 - 32)^2 + (34 - 32)^2 + (34 - 32)^2 + (31 - 32)^2}{6}\]\[= \frac{9 + 0 + 0 + 4 + 4 + 1}{6} = \frac{18}{6} = 3\]

06

Calculate the Standard Deviation for Both Samples

The standard deviation is the square root of the variance.For cities:\[\text{Standard Deviation} = \sqrt{12} \approx 3.46\]For retirement areas:\[\text{Standard Deviation} = \sqrt{3} \approx 1.73\]

07

Analyze the Results

The mean grocery costs are higher in cities ($33) compared to retirement areas ($32). The variance and standard deviation are also higher in cities, indicating more fluctuation in grocery costs compared to retirement areas.

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Key Concepts

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

Mean Calculation

In statistics, the mean gives us an average value, which helps to understand the central tendency of a dataset. The mean is calculated by adding up all sample values and dividing by the number of samples.

For cities in the original exercise, the grocery costs per city are

• Buffalo, NY: \(33
• Des Moines, IA: \)27
• Hartford, CT: \(32
• Los Angeles, CA: \)38
• Miami, FL: \(36
• Pittsburgh, PA: \)32

To calculate the mean cost for cities, sum all the costs and divide by the number of cities: \[\text{Mean for cities} = \frac{33 + 27 + 32 + 38 + 36 + 32}{6} = \frac{198}{6} = 33 \] Similarly, for the retirement areas with costs

• Biloxi-Gulfport, MS: \(29
• Asheville, NC: \)32
• Flagstaff, AZ: \(32
• Hilton Head, SC: \)34
• Fort Myers, FL: \(34
• Santa Fe, NM: \)31

The mean cost is \[\text{Mean for retirement areas} = \frac{29 + 32 + 32 + 34 + 34 + 31}{6} = \frac{192}{6} = 32 \] This tells us that, on average, grocery costs are slightly lower in retirement areas compared to cities.

Variance Calculation

Variance is a measure that indicates how much the values in a dataset are spread out. A higher variance means data points are more dispersed from the mean.

For this analysis, we calculate the variance by taking each value from the dataset, subtracting the mean, squaring the result, summing these squared differences, and dividing by the number of samples.
For the city costs:

• Mean: 33
• Values: 33, 27, 32, 38, 36, 32

Variance formula for cities looks like this: \[\text{Variance for cities} = \frac{(33 - 33)^2 + (27 - 33)^2 + (32 - 33)^2 + (38 - 33)^2 + (36 - 33)^2 + (32 - 33)^2}{6}\]\[= \frac{0 + 36 + 1 + 25 + 9 + 1}{6} = \frac{72}{6} = 12 \] For the retirement areas, with mean 32 and costs as 29, 32, 32, 34, 34, 31, the variance is: \[\text{Variance for retirement areas} = \frac{(29 - 32)^2 + (32 - 32)^2 + (32 - 32)^2 + (34 - 32)^2 + (34 - 32)^2 + (31 - 32)^2}{6} \] \[= \frac{9 + 0 + 0 + 4 + 4 + 1}{6} = \frac{18}{6} = 3 \] The cities have a higher variance, showing more variability in grocery costs compared to retirement areas.

Standard Deviation

Standard deviation is a useful statistic for quantifying the amount of variation or dispersion in a dataset. It is a measure of how spread out numbers are, calculated as the square root of the variance.
A larger standard deviation indicates a wider diversity in data values.
Here, we calculate the standard deviation for the cities' and retirement areas' market basket costs. For the cities: the variance is 12. The standard deviation is calculated by taking the square root of the variance: \[\text{Standard Deviation for cities} = \sqrt{12} \approx 3.46 \] This value suggests a moderate variation in grocery costs across different cities.
For the retirement areas: with a variance of 3, the standard deviation is \[\text{Standard Deviation for retirement areas} = \sqrt{3} \approx 1.73 \] This lower value indicates less variability in costs compared to the cities.
Overall, standard deviation provides insight into how spread out the costs are around the mean, helping us grasp how consistent the spending is across different locations.