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Chapter 6: Q7E (page 262)

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

Expert verified

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

01

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.

02

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 \({\rm{120}}{\rm{.6}}\).

03

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 \({\rm{1,206,000}}\).

04

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 \({\rm{0}}{\rm{.8}}\).

05

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 \({\rm{120}}\)

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Most popular questions from this chapter

Suppose the true average growth\({\rm{\mu }}\)of one type of plant during a l-year period is identical to that of a second type, but the variance of growth for the first type is\({{\rm{\sigma }}^{\rm{2}}}\), whereas for the second type the variance is\({\rm{4}}{{\rm{\sigma }}^{\rm{2}}}{\rm{. Let }}{{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{m}}}\)be\({\rm{m}}\)independent growth observations on the first type (so\({\rm{E}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ = \mu ,V}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ = \sigma\hat 2}}\)$ ), and let\({{\rm{Y}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{Y}}_{\rm{n}}}\)be\({\rm{n}}\)independent growth observations on the second type\(\left( {{\rm{E}}\left( {{{\rm{Y}}_{\rm{i}}}} \right){\rm{ = \mu ,V}}\left( {{{\rm{Y}}_{\rm{j}}}} \right){\rm{ = 4}}{{\rm{\sigma }}^{\rm{2}}}} \right)\)

a. Show that the estimator\({\rm{\hat \mu = \delta \bar X + (1 - \delta )\bar Y}}\)is unbiased for\({\rm{\mu }}\)(for\({\rm{0 < \delta < 1}}\), the estimator is a weighted average of the two individual sample means).

b. For fixed\({\rm{m}}\)and\({\rm{n}}\), compute\({\rm{V(\hat \mu ),}}\)and then find the value of\({\rm{\delta }}\)that minimizes\({\rm{V(\hat \mu )}}\). (Hint: Differentiate\({\rm{V(\hat \mu )}}\)with respect to\({\rm{\delta }}{\rm{.)}}\)

At time \({\rm{t = 0}}\), there is one individual alive in a certain population. A pure birth process then unfolds as follows. The time until the first birth is exponentially distributed with parameter \({\rm{\lambda }}\). After the first birth, there are two individuals alive. The time until the first gives birth again is exponential with parameter \({\rm{\lambda }}\), and similarly for the second individual. Therefore, the time until the next birth is the minimum of two exponential (\({\rm{\lambda }}\)) variables, which is exponential with parameter \({\rm{2\lambda }}\). Similarly, once the second birth has occurred, there are three individuals alive, so the time until the next birth is an exponential \({\rm{rv}}\) with parameter \({\rm{3\lambda }}\), and so on (the memoryless property of the exponential distribution is being used here). Suppose the process is observed until the sixth birth has occurred and the successive birth times are \({\rm{25}}{\rm{.2,41}}{\rm{.7,51}}{\rm{.2,55}}{\rm{.5,59}}{\rm{.5,61}}{\rm{.8}}\) (from which you should calculate the times between successive births). Derive the mle of l. (Hint: The likelihood is a product of exponential terms.)

The mean squared error of an estimator \({\rm{\hat \theta }}\) is \({\rm{MSE(\hat \theta ) = E(\hat \theta - \hat \theta }}{{\rm{)}}^{\rm{2}}}\). If \({\rm{\hat \theta }}\) is unbiased, then \({\rm{MSE(\hat \theta ) = V(\hat \theta )}}\), but in general \({\rm{MSE(\hat \theta ) = V(\hat \theta ) + (bias}}{{\rm{)}}^{\rm{2}}}\) . Consider the estimator \({{\rm{\hat \sigma }}^{\rm{2}}}{\rm{ = K}}{{\rm{S}}^{\rm{2}}}\), where \({{\rm{S}}^{\rm{2}}}{\rm{ = }}\) sample variance. What value of K minimizes the mean squared error of this estimator when the population distribution is normal? (Hint: It can be shown that \({\rm{E}}\left( {{{\left( {{{\rm{S}}^{\rm{2}}}} \right)}^{\rm{2}}}} \right){\rm{ = (n + 1)}}{{\rm{\sigma }}^{\rm{4}}}{\rm{/(n - 1)}}\) In general, it is difficult to find \({\rm{\hat \theta }}\) to minimize \({\rm{MSE(\hat \theta )}}\), which is why we look only at unbiased estimators and minimize \({\rm{V(\hat \theta )}}\).)

The accompanying data on flexural strength (MPa) for concrete beams of a certain type was introduced in Example 1.2.

\(\begin{array}{*{20}{r}}{{\rm{5}}{\rm{.9}}}&{{\rm{7}}{\rm{.2}}}&{{\rm{7}}{\rm{.3}}}&{{\rm{6}}{\rm{.3}}}&{{\rm{8}}{\rm{.1}}}&{{\rm{6}}{\rm{.8}}}&{{\rm{7}}{\rm{.0}}}\\{{\rm{7}}{\rm{.6}}}&{{\rm{6}}{\rm{.8}}}&{{\rm{6}}{\rm{.5}}}&{{\rm{7}}{\rm{.0}}}&{{\rm{6}}{\rm{.3}}}&{{\rm{7}}{\rm{.9}}}&{{\rm{9}}{\rm{.0}}}\\{{\rm{3}}{\rm{.2}}}&{{\rm{8}}{\rm{.7}}}&{{\rm{7}}{\rm{.8}}}&{{\rm{9}}{\rm{.7}}}&{{\rm{7}}{\rm{.4}}}&{{\rm{7}}{\rm{.7}}}&{{\rm{9}}{\rm{.7}}}\\{{\rm{7}}{\rm{.3}}}&{{\rm{7}}{\rm{.7}}}&{{\rm{11}}{\rm{.6}}}&{{\rm{11}}{\rm{.3}}}&{{\rm{11}}{\rm{.8}}}&{{\rm{10}}{\rm{.7}}}&{}\end{array}\)

Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion, and state which estimator you used\({\rm{(Hint:\Sigma }}{{\rm{x}}_{\rm{i}}}{\rm{ = 219}}{\rm{.8}}{\rm{.)}}\)

b. Calculate a point estimate of the strength value that separates the weakest 50% of all such beams from the strongest 50 %, and state which estimator you used.

c. Calculate and interpret a point estimate of the population standard deviation\({\rm{\sigma }}\). Which estimator did you use?\({\rm{(Hint:}}\left. {{\rm{\Sigma x}}_{\rm{i}}^{\rm{2}}{\rm{ = 1860}}{\rm{.94}}{\rm{.}}} \right)\)

d. Calculate a point estimate of the proportion of all such beams whose flexural strength exceeds\({\rm{10MPa}}\). (Hint: Think of an observation as a "success" if it exceeds 10.)

e. Calculate a point estimate of the population coefficient of variation\({\rm{\sigma /\mu }}\), and state which estimator you used.

Let\({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\)represent a random sample from a Rayleigh distribution with pdf

\({\rm{f(x,\theta ) = }}\frac{{\rm{x}}}{{\rm{\theta }}}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/(2\theta )}}}}\quad {\rm{x > 0}}\)a. It can be shown that\({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ = 2\theta }}\). Use this fact to construct an unbiased estimator of\({\rm{\theta }}\)based on\({\rm{\Sigma X}}_{\rm{i}}^{\rm{2}}\)(and use rules of expected value to show that it is unbiased).

b. Estimate\({\rm{\theta }}\)from the following\({\rm{n = 10}}\)observations on vibratory stress of a turbine blade under specified conditions:

\(\begin{array}{*{20}{l}}{{\rm{16}}{\rm{.88}}}&{{\rm{10}}{\rm{.23}}}&{{\rm{4}}{\rm{.59}}}&{{\rm{6}}{\rm{.66}}}&{{\rm{13}}{\rm{.68}}}\\{{\rm{14}}{\rm{.23}}}&{{\rm{19}}{\rm{.87}}}&{{\rm{9}}{\rm{.40}}}&{{\rm{6}}{\rm{.51}}}&{{\rm{10}}{\rm{.95}}}\end{array}\)
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