Question

A random sample of 10 houses in a particular area, each of which is heated with natural gas, is selected, and the amount of gas (in therms) used during the month of January is determined for each house. The resulting observations are as follows: \(\begin{array}{lllllllll}103 & 156 & 118 & 89 & 125 & 147 & 122 & 109 & 138 & 99\end{array}\) a. Let \(\mu_{j}\) denote the average gas usage during January by all houses in this area. Compute a point estimate of \(\mu_{J}\). b. Suppose that 10,000 houses in this area use natural gas for heating. Let \(\tau\) denote the total amount of gas used by all of these houses during January. Estimate \(\tau\) using the data of Part (a). What statistic did you use in computing your estimate? c. Use the data in Part (a) to estimate \(\pi\), the proportion of all houses that used at least 100 therms. d. Give a point estimate of the population median usage based on the sample of Part (a). Which statistic did you use?

Short answer

a. Average gas usage (point estimate of \( \mu_J \)) = 120.6 therms. b. Total gas usage estimate (\( \tau \)) = 1,206,000 therms. c. Proportion of houses that used at least 100 therms (\( \pi \)) = 90% d. Population median usage point estimate = 122.5 therms

Step-by-step solution

Compute a Point Estimate of Average Gas Usage

To estimate the average gas usage during January by all houses in this area, represented as \( \mu_J \), we will use the mean of the observed gas usages for the 10 randomly chosen houses. The average is computed as follows: \[\mu_{j} = \frac{1}{n}\Sigma_{i=1}^{n} x_i \], where \(x_i\) represents individual observations and \(n\) is the number of observations.

Estimate Total Gas Usage

To estimate total gas usage, represented as \( \tau \), we multiply the average gas usage per house, \( \mu_j \), with the total number of houses in the area (10,000), i.e. \( \tau = \mu_j * total houses \) . The statistic used for computing this is the average computed in Step 1.

Estimate Proportion of Houses that Used at Least 100 Therms

To estimate \( \pi \), the proportion of all houses that used at least 100 therms, we compute the proportion of the selected houses that used at least 100 therms. That is \( \pi = \frac{houses that used ≥ 100 therms}{total houses (from the sample)} \) .

Point Estimate of Population Median

To give a point estimate of the population median usage, we will use the median of the observed usages for the 10 randomly chosen houses. The median is computed as follows: if \( n \) is odd, the median is the value at position \( (n+1)/2 \) after sorting the data in ascending order. If \( n \) is even, the median is the mean of the values at positions \( n/2 \) and \( n/2 + 1 \). The statistic used is the median of the sample data.

Key concepts

Average Gas Usage

Understanding average gas usage is crucial for making informed decisions about energy consumption and budgeting. The average gas usage refers to the mean amount of gas consumed over a certain period. In our textbook problem, we look at a sample of 10 houses to find the average gas usage during January represented by \( \mu_J \). To compute this value, we sum up the gas usage in therms for all houses and then divide by the number of observations, following the formula: \[ \mu_{j} = \frac{1}{n}\Sigma_{i=1}^{n} x_i \], where \(x_i\) is the usage for each house, and \(n\) is the sample size.
Calculating this point estimate provides insight into typical consumption patterns and establishes a basis for comparison and estimation of larger populations. This average can help utility companies to forecast demand and assist in the planning of the gas supply.

Total Gas Usage Estimation

Estimating the total gas usage for a larger population based on a sample can be incredibly useful for planning and resource allocation. After finding the average gas usage, we can estimate the total usage, denoted as \(\tau\), for a given number of households. In the provided exercise, we estimated the total gas usage for 10,000 houses by multiplying the point estimate of the average usage \( \mu_J \) by the number of houses.

The equation is straightforward: \( \tau = \mu_j \times \text{total houses} \). The average gas usage acts as a representative statistic that assumes each house in the population consumes gas at the same rate as our sample. This simplification can help utility companies predict total consumption and ensure they have sufficient supply for all their customers during high-demand periods like January.

Proportion Estimation

Proportion estimation in statistics is useful for determining the percentage of a population that exhibits a particular characteristic. In the context of our textbook problem, we use it to estimate the proportion of houses, symbolized by \(\pi\), that used at least 100 therms of gas during January. To calculate this proportion, we count the number of houses in our sample that meet the criterion and divide that by the total sample size. The formula looks like this: \( \pi = \frac{\text{houses that used} \geq 100 \text{ therms}}{\text{total houses (from the sample)}} \).

This estimation helps utility companies to understand customer usage patterns and can serve as a basis to categorize households into different consumption tiers.

Population Median Estimation

The median is a measure of central tendency that indicates the middle value of a dataset when the values are arranged in ascending order. Estimating the population median based on a sample is an exercise in finding which usage value represents the midpoint of all households' usage. Unlike the mean, the median is not influenced by extreme values and may provide a better sense of 'typical' usage in a skewed distribution.

In the solved problem, we sorted the gas usage of the 10 houses and identified the median, as our sample size was even, the median is the average of the 5th and 6th values after sorting the data. The formula for an even number of observations \(n\) is: if \( n \) is even, the median is the mean of the values at positions \( n/2 \) and \( n/2 + 1 \). This point estimate of the population median provides a robust central value which, unlike the mean, is not distorted by outliers.