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A company maintains three offices in a certain region, each staffed by two employees. Information concerning yearly salaries (\({\rm{1000}}\)s of dollars) is as follows:

\(\begin{array}{*{20}{c}}{{\rm{ Office }}}&{\rm{1}}&{\rm{1}}&{\rm{2}}&{\rm{2}}&{\rm{3}}&{\rm{3}}\\{{\rm{ Employee }}}&{\rm{1}}&{\rm{2}}&{\rm{3}}&{\rm{4}}&{\rm{5}}&{\rm{6}}\\{{\rm{ Salary }}}&{{\rm{29}}{\rm{.7}}}&{{\rm{33}}{\rm{.6}}}&{{\rm{30}}{\rm{.2}}}&{{\rm{33}}{\rm{.6}}}&{{\rm{25}}{\rm{.8}}}&{{\rm{29}}{\rm{.7}}}\end{array}\)

a. Suppose two of these employees are randomly selected from among the six (without replacement). Determine the sampling distribution of the sample mean salary\({\rm{\bar X}}\).

b. Suppose one of the three offices is randomly selected. Let\({{\rm{X}}_{\rm{1}}}\)and\({{\rm{X}}_{\rm{2}}}\)denote the salaries of the two employees. Determine the sampling distribution of\({\rm{\bar X}}\).

c. How does \({\rm{E(\bar X)}}\) from parts (a) and (b) compare to the population mean salary\({\rm{\mu }}\)?

Short Answer

Expert verified

a) The sampling distribution

b) The sampling distribution

c) The population mean salary \({\rm{\mu }}\) is equal to the anticipated values \({\rm{E(\bar X)}}\) from parts (a) and (b).

Step by step solution

01

Definition

Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes—how likely they are—when we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

02

Determine the sampling distribution of the sample mean salary \({\rm{\bar X}}\)

Determine the sample mean salary (sum of all values divided by the number of values) for each sample of \({\rm{2}}\) employees:

Determine the number of times each sample mean appears and the probability associated with it (frequency divided by the total frequency of \({\rm{30}}\)).

03

Determining the sampling distribution of \({\rm{\bar X}}\)

b. Calculate the sample mean pay for each sample comprising one office (sum of all values divided by the number of values):


Determine the number of times each sample mean appears and the probability associated with it (frequency divided by the total frequency of \({\rm{3}}\)).

04

Determine the sampling distribution of the sample mean salary \({\rm{\bar X}}\)

c. The anticipated value is calculated by multiplying each alternative by its probability:

\(\begin{aligned}{\rm{E}}\left( {{{{\rm{\bar X}}}_{\rm{a}}}} \right)\rm &= \sum {\rm{x}} {\rm{P(x)}}\\\rm &= 27{\rm{.75}}\frac{{\rm{4}}}{{{\rm{30}}}}{\rm{ + 28}}\frac{{\rm{2}}}{{{\rm{30}}}}{\rm{ + 29}}{\rm{.7}}\frac{{\rm{6}}}{{{\rm{30}}}}{\rm{ + 29}}{\rm{.95}}\frac{{\rm{4}}}{{{\rm{30}}}}{\rm{ + 31}}{\rm{.65}}\frac{{\rm{8}}}{{{\rm{30}}}}{\rm{ + 31}}{\rm{.9}}\frac{{\rm{4}}}{{{\rm{30}}}}{\rm{ + 33}}{\rm{.6}}\frac{{\rm{2}}}{{{\rm{30}}}}\\&{\rm{\gg 30}}{\rm{.4333}}\\{\rm{E}}\left( {{{{\rm{\bar X}}}_{\rm{b}}}} \right)&{\rm{ = }}\sum {\rm{x}} {\rm{P(x)}}\\&{\rm{ = 27}}{\rm{.75}}\frac{{\rm{1}}}{{\rm{3}}}{\rm{ + 31}}{\rm{.65}}\frac{{\rm{1}}}{{\rm{3}}}{\rm{ + 31}}{\rm{.9}}\frac{{\rm{1}}}{{\rm{3}}}\\&{\rm{\gg 30}}{\rm{.4333}}\end{aligned}\)

The population mean is the sum of all values divided by the number of values:

\({\rm{\mu = }}\frac{{{\rm{29}}{\rm{.7 + 33}}{\rm{.6 + 30}}{\rm{.2 + 33}}{\rm{.6 + 25}}{\rm{.8 + 29}}{\rm{.7}}}}{{\rm{6}}}{\rm{\gg 30}}{\rm{.4333}}\)

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