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Business / Statistics for Business and Economics 12 / Chapter 6 / Problem 86E

Statistics for Business and Economics | 12th Edition | ISBN: 9780321826237 | Authors: James T. McClave, P. George Benson, Terry T Sincich

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Magazine subscriber salaries. Each year, the trade

Chapter 6, Problem 86E

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QUESTION:

Problem 86E

Magazine subscriber salaries. Each year, the trade magazine Quality Progress publishes a study of subscribers’ salaries. One year, the 223 vice presidents sampled had a mean salary of $116,754 and a standard deviation of $39,185. Suppose the goal of the study is to estimate the true mean salary of all vice presidents who subscribe to Quality Progress.

a. If 2,193 vice presidents subscribe to Quality Progress, estimate the mean with an approximate 95% confidence interval.

b. Interpret the result.

Questions & Answers

QUESTION:

Problem 86E

Magazine subscriber salaries. Each year, the trade magazine Quality Progress publishes a study of subscribers’ salaries. One year, the 223 vice presidents sampled had a mean salary of $116,754 and a standard deviation of $39,185. Suppose the goal of the study is to estimate the true mean salary of all vice presidents who subscribe to Quality Progress.

a. If 2,193 vice presidents subscribe to Quality Progress, estimate the mean with an approximate 95% confidence interval.

b. Interpret the result.

ANSWER:

Answer

Step 1 of 2

(a)

One year, the 223 vice presidents sampled had a mean salary of $116,754 and a standard deviation of $39,185.

Suppose the goal of the study is to estimate the true mean salary of all vice presidents who subscribe to quality progress.

We are asked to find the estimate of mean with an approximate 95% confidence interval if 2,193 vice presidents subscribe to quality progress.

From the Empirical Rule, if we assume population comes from the normal population, then we can write 95% confidence interval as $\overline{x}\ \pm{}2{\sigma{}}_{x}.$

Finite population correction factor: The term $\left({\sqrt{\frac{N\ -n}{N-1}}}\right)$ that is used in the formulas for ${\sigma{}}_{\overline{x}}$ and ${\sigma{}}_{\overline{p}}$ whenever a finite population, rather than an infinite population, is being sampled.

The generally accepted rule of thumb is to ignore the finite population correction factor whenever $\frac{n}{N}\leq{}0.050.$

Since,

$\frac{n}{N}\ =\frac{233}{2193}\ =0.10$

We need to use the population correction factor because of $\frac{n}{N}>0.050.$

Hence for finite population, standard deviation of $\overline{x,}$

${\sigma{}}_{x}\ =\frac{s}{\sqrt{n}}\left({\sqrt{\frac{N\ -n}{N-1}}}\right)$

Hence a level $95\%$ confidence interval for mean ${\mu{}}_{X}\$ ( interval estimate) is,

$\overline{x}\ \pm{}2{\sigma{}}_{x}$

$\overline{x}\ \pm{}2\times{}\frac{s}{\sqrt{n}}\left({\sqrt{\frac{N\ -n}{N-1}}}\right)$

We have given $N\ =2,193,\ n\ =223,\ \overline{x}\ =\$116,754,\ and\ s\ =\$39,185.$

$116,754\ \pm{}2\times{}\frac{39,185}{\sqrt{223}}\left({\sqrt{\frac{2193\ -223}{2193-1}}}\right)$

$(\$116,754\ \pm{}4975.2)$

Hence the estimate of mean with an approximate 95% confidence interval is $(\$111,778.8,\ \ \$121,729.2).$

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