Question

a. 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 (therms) used during the month of January is determined for each house. The resulting observations are 103, 156, 118, 89, 125, 147, 122, 109, 138, 99. Let m denote the average gas usage during January by all houses in this area. Compute a point estimate of m. b. Suppose there are 10,000 houses in this area that use natural gas for heating. Let t denote the total amount of gas used by all of these houses during January. Estimate t using the data of part (a). What estimator did you use in computing your estimate? c. Use the data in part (a) to estimate p, the proportion of all houses that used at least 100 therms. d. Give a point estimate of the population median usage (the middle value in the population of all houses) based on the sample of part (a). What estimator did you use?

Short answer

a.The mean value is \({\rm{120}}{\rm{.6}}\).

b.Using the estimate from (a) for \(\mu \)we get \({\rm{1,206,000}}\).

c.The estimate of the proportion is \({\rm{0}}{\rm{.8}}\).

d.The sample median is average of the two mentioned values, is \ ({\rm{120}}\).

Step-by-step solution

Concept introduction

The standard deviation (SD) is a measure of the variability, or dispersion, between individual data values and the mean, whereas the standard error of the mean (SEM) is a measure of how far the sample mean (average) of the data is expected to differ from the genuine population mean. Always, the SEM is smaller than the SD.

Observing the sample mean

a.

The Sample Mean \(\bar x\) of observations \({x_1},{x_2}, \ldots ,{x_n}\)is given by

\(\begin{array}{c}\bar x = \frac{{{x_1} + {x_2} + \ldots + {x_n}}}{n}\\ = \frac{1}{n}\sum\limits_{i = 1}^n {{x_i}} \end{array}\)

The sample mean can be used for the point estimate of the mean value \((\mu )\)

\(\begin{array}{c}{\rm{\bar x = }}\frac{{{{\rm{x}}_{\rm{1}}}{\rm{ + }}{{\rm{x}}_{\rm{2}}}{\rm{ + \ldots + }}{{\rm{x}}_{\rm{n}}}}}{{\rm{n}}}\\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{10}}}}{\rm{(103 + 156 + \ldots + 99)}}\\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{10}}}}{\rm{ \times 1206}}\\{\rm{ = 120}}{\rm{.6}}\end{array}\)

Hence, the mean value is \({\rm{120}}{\rm{.6}}\).

Calculating part (b)

(b)

There are \({\rm{N = 10,000}}\)residences in the neighborhood. The total amount of gas used, \(\tau \), may be calculated with the help of the estimator .

\({\rm{\hat \tau = N \times \hat \mu }}\)

Using the estimate from (a) for \(\mu \)we get

\(\begin{array}{c}{\rm{\hat \tau = 10,000 \times 120}}{\rm{.6}}\\{\rm{ = 1,206,000}}\end{array}\)

Hence, using the estimate from (a) for \(\mu \)we get \({\rm{1,206,000}}\).

Finding estimate proportion

(c)

Because eight out of ten values in the given data are less than one hundred, the estimate of the proportion of all households that used at least one hundred terms is

\(\begin{array}{c}{\rm{\hat p = }}\frac{{\rm{8}}}{{{\rm{10}}}}\\{\rm{ = 0}}{\rm{.8}}\end{array}\)

Thus, the estimate of the proportion is \({\rm{0}}{\rm{.8}}\).

Estimating sample Median

(d)

The sample median can be used to get the median point estimate.

To begin, arrange the data in the following order:

\({\rm{89,99,103,109,118,122,125,138,147,156 }}\)

Because this sample has an even number of \({\rm{n = 10}}\)observations, the values of interest are.

\(\begin{array}{c}{\left( {\frac{{\rm{n}}}{{\rm{2}}}} \right)^{{\rm{th}}}}{\rm{ = }}{\left( {\frac{{{\rm{10}}}}{{\rm{2}}}} \right)^{{\rm{th}}}}\\{\rm{ = }}{{\rm{5}}^{{\rm{th}}}}\end{array}\)

where the \({5^{th}}\)and \({6^{th}}\)values are marked red in the ordered sample data. The sample median is average of the two mentioned values, hence

\(\begin{array}{c}{\rm{\tilde x = }}\frac{{{\rm{118 + 122}}}}{{\rm{2}}}\\{\rm{ = 120}}\end{array}\)

Therefore, the sample median is average of the two mentioned values, is \({\rm{120}}\)