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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

The article from which the data in Exercise 1 was extracted also gave the accompanying strength observations for cylinders:

\(\begin{array}{l}\begin{array}{*{20}{r}}{{\rm{6}}{\rm{.1}}}&{{\rm{5}}{\rm{.8}}}&{{\rm{7}}{\rm{.8}}}&{{\rm{7}}{\rm{.1}}}&{{\rm{7}}{\rm{.2}}}&{{\rm{9}}{\rm{.2}}}&{{\rm{6}}{\rm{.6}}}&{{\rm{8}}{\rm{.3}}}&{{\rm{7}}{\rm{.0}}}&{{\rm{8}}{\rm{.3}}}\\{{\rm{7}}{\rm{.8}}}&{{\rm{8}}{\rm{.1}}}&{{\rm{7}}{\rm{.4}}}&{{\rm{8}}{\rm{.5}}}&{{\rm{8}}{\rm{.9}}}&{{\rm{9}}{\rm{.8}}}&{{\rm{9}}{\rm{.7}}}&{{\rm{14}}{\rm{.1}}}&{{\rm{12}}{\rm{.6}}}&{{\rm{11}}{\rm{.2}}}\end{array}\\\begin{array}{*{20}{l}}{{\rm{7}}{\rm{.8}}}&{{\rm{8}}{\rm{.1}}}&{{\rm{7}}{\rm{.4}}}&{{\rm{8}}{\rm{.5}}}&{{\rm{8}}{\rm{.9}}}&{{\rm{9}}{\rm{.8}}}&{{\rm{9}}{\rm{.7}}}&{{\rm{14}}{\rm{.1}}}&{{\rm{12}}{\rm{.6}}}&{{\rm{11}}{\rm{.2}}}\end{array}\end{array}\)

Prior to obtaining data, denote the beam strengths by X1, 鈥� ,Xm and the cylinder strengths by Y1, . . . , Yn. Suppose that the Xi 鈥檚 constitute a random sample from a distribution with mean m1 and standard deviation s1 and that the Yi 鈥檚 form a random sample (independent of the Xi 鈥檚) from another distribution with mean m2 and standard deviation\({{\rm{\sigma }}_{\rm{2}}}\).

a. Use rules of expected value to show that \({\rm{\bar X - \bar Y}}\)is an unbiased estimator of \({{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}\). Calculate the estimate for the given data.

b. Use rules of variance from Chapter 5 to obtain an expression for the variance and standard deviation (standard error) of the estimator in part (a), and then compute the estimated standard error.

c. Calculate a point estimate of the ratio \({{\rm{\sigma }}_{\rm{1}}}{\rm{/}}{{\rm{\sigma }}_{\rm{2}}}\)of the two standard deviations.

d. Suppose a single beam and a single cylinder are randomly selected. Calculate a point estimate of the variance of the difference \({\rm{X - Y}}\) between beam strength and cylinder strength.

Urinary angiotensinogen (AGT) level is one quantitative indicator of kidney function. The article 鈥淯rinary Angiotensinogen as a Potential Biomarker of Chronic Kidney Diseases鈥� (J. of the Amer. Society of Hypertension, \({\rm{2008: 349 - 354}}\)) describes a study in which urinary AGT level \({\rm{(\mu g)}}\) was determined for a sample of adults with chronic kidney disease. Here is representative data (consistent with summary quantities and descriptions in the cited article):

An appropriate probability plot supports the use of the lognormal distribution (see Section \({\rm{4}}{\rm{.5}}\)) as a reasonable model for urinary AGT level (this is what the investigators did).

a. Estimate the parameters of the distribution. (Hint: Rem ember that \({\rm{X}}\) has a lognormal distribution with parameters \({\rm{\mu }}\) and \({{\rm{\sigma }}^{\rm{2}}}\) if \({\rm{ln(X)}}\) is normally distributed with mean \({\rm{\mu }}\) and variance \({{\rm{\sigma }}^{\rm{2}}}\).)

b. Use the estimates of part (a) to calculate an estimate of the expected value of AGT level. (Hint: What is \({\rm{E(X)}}\)?)

An estimator \({\rm{\hat \theta }}\) is said to be consistent if for any \( \in {\rm{ > 0}}\), \({\rm{P(|\hat \theta - \theta |}} \ge \in {\rm{)}} \to {\rm{0}}\) as \({\rm{n}} \to \infty \). That is, \({\rm{\hat \theta }}\) is consistent if, as the sample size gets larger, it is less and less likely that \({\rm{\hat \theta }}\) will be further than \( \in \) from the true value of \({\rm{\theta }}\). Show that \({\rm{\bar X}}\) is a consistent estimator of \({\rm{\mu }}\) when \({{\rm{\sigma }}^{\rm{2}}}{\rm{ < }}\infty \) , by using Chebyshev鈥檚 inequality from Exercise \({\rm{44}}\) of Chapter \({\rm{3}}\). (Hint: The inequality can be rewritten in the form \({\rm{P}}\left( {\left| {{\rm{Y - }}{{\rm{\mu }}_{\rm{Y}}}} \right| \ge \in } \right) \le {\rm{\sigma }}_{\rm{Y}}^{\rm{2}}{\rm{/}} \in \). Now identify \({\rm{Y}}\) with \({\rm{\bar X}}\).)

Let\({\rm{X}}\)have a Weibull distribution with parameters\({\rm{\alpha }}\)and\({\rm{\beta }}\), so

\(\begin{array}{l}{\rm{E(X) = \beta \times \Gamma (1 + 1/\alpha )V(X)}}\\{\rm{ = }}{{\rm{\beta }}^{\rm{2}}}\left\{ {{\rm{\Gamma (1 + 2/\alpha ) - (\Gamma (1 + 1/\alpha )}}{{\rm{)}}^{\rm{2}}}} \right\}\end{array}\)

a. Based on a random sample\({{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\), write equations for the method of moments estimators of\({\rm{\beta }}\)and\({\rm{\alpha }}\). Show that, once the estimate of\({\rm{\alpha }}\)has been obtained, the estimate of\({\rm{\beta }}\)can be found from a table of the gamma function and that the estimate of\({\rm{\alpha }}\)is the solution to a complicated equation involving the gamma function.

b. If\({\rm{n = 20,\bar x = 28}}{\rm{.0}}\), and\({\rm{\Sigma x}}_{\rm{i}}^{\rm{2}}{\rm{ = 16,500}}\), compute the estimates. (Hint:\(\left. {{{{\rm{(\Gamma (1}}{\rm{.2))}}}^{\rm{2}}}{\rm{/\Gamma (1}}{\rm{.4) = }}{\rm{.95}}{\rm{.}}} \right)\)

Consider a random sample \({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}.....{\rm{,}}{{\rm{X}}_{\rm{n}}}\) from the shifted exponential pdf

\({\rm{f(x;\lambda ,\theta ) = }}\left\{ {\begin{array}{*{20}{c}}{{\rm{\lambda }}{{\rm{e}}^{{\rm{ - \lambda (x - \theta )}}}}}&{{\rm{x}} \ge {\rm{\theta }}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\). Taking \({\rm{\theta = 0}}\) gives the pdf of the exponential distribution considered previously (with positive density to the right of zero). An example of the shifted exponential distribution appeared in Example \({\rm{4}}{\rm{.5}}\), in which the variable of interest was time headway in traffic flow and \({\rm{\theta = }}{\rm{.5}}\) was the minimum possible time headway. a. Obtain the maximum likelihood estimators of \({\rm{\theta }}\) and \({\rm{\lambda }}\). b. If \({\rm{n = 10}}\) time headway observations are made, resulting in the values \({\rm{3}}{\rm{.11,}}{\rm{.64,2}}{\rm{.55,2}}{\rm{.20,5}}{\rm{.44,3}}{\rm{.42,10}}{\rm{.39,8}}{\rm{.93,17}}{\rm{.82}}\), and \({\rm{1}}{\rm{.30}}\), calculate the estimates of \({\rm{\theta }}\) and \({\rm{\lambda }}\).
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