Question

Annual starting salaries of college graduates with degrees in business administration are generally expected to be between $30,000 and $45,000.

1) Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired. Determine the planning value for the population standarddeviation.
2) Determine how large a sample should be taken if the desired margin of error is:
a. $500
b. $200
c. $100
d. Would you recommend trying to obtain the $100 margin of error? Explain

Answer 1

Concepts and reason

The concept is based on planning value for the standard deviation and sample size for desired margin of error.

The planning value of the standard deviation of a sample is approximately equal to one fourth of the range of the data. It provides only a very rough estimate of the standard deviation.

Sample size is the size of the provided sample and margin of error is the width of the confidence interval for the statistic.

Fundamentals

The planning value for the population standard deviation is provided by,

$s = \frac{{{\rm{Maximum}} - {\rm{Minimum}}}}{4}$

The formula to calculate the margin of error is,

$n = {\left( {\frac{{z\sigma }}{E}} \right)^2}$

(1)

Assume that the confidence level is 95%. The maximum value of the annual starting salaries of college graduates is provided as $45,000 and the minimum value is provided as $30,000.

The planning value for population standard deviation is,

$\begin{array}{c}\\s = \frac{{{\rm{Maximum}} - {\rm{Minimum}}}}{4}\\\\ = \frac{{45000 - 30000}}{4}\\\\ = \frac{{15000}}{4}\\\end{array}$

$= 3750$

(2.a)

The standard deviation is obtained as $3,750 in the previous part. Provided that the margin of error is $500. Since the confidence level is 95%, $\alpha = 0.05$ and the Z – score corresponding to this confidence level is 1.96 from the standard normal table.

Substitute these values to compute the sample size as,

$\begin{array}{c}\\n = {\left( {\frac{{z\sigma }}{E}} \right)^2}\\\\ = {\left( {\frac{{\left( {1.96} \right)\left( {3750} \right)}}{{500}}} \right)^2}\\\\ = {\left( {14.7} \right)^2}\\\end{array}$

$\begin{array}{l}\\ = 216.09\\\\ \approx 216\\\end{array}$

(2.b)

The standard deviation is obtained as $3,750 in the previous part. Provided that the margin of error is $200. Since the confidence level is 95%, $\alpha = 0.05$ and the Z – score corresponding to this confidence level is 1.96 from the standard normal table.

Substitute these values to compute the sample size as,

$\begin{array}{c}\\n = {\left( {\frac{{z\sigma }}{E}} \right)^2}\\\\ = {\left( {\frac{{\left( {1.96} \right)\left( {3750} \right)}}{{200}}} \right)^2}\\\\ = {36.75^2}\\\end{array}$

$\begin{array}{l}\\ = 1350.56\\\\ \approx 1351\\\end{array}$

(2.c)

The standard deviation is obtained as $3,750 in the previous part. Provided that the margin of error is $100. Since the confidence level is 95%, $\alpha = 0.05$ and the Z – score corresponding to this confidence level is 1.96 from the standard normal table.

Substitute these values to compute the sample size as,

$\begin{array}{c}\\n = {\left( {\frac{{z\sigma }}{E}} \right)^2}\\\\ = {\left( {\frac{{\left( {1.96} \right)\left( {3,750} \right)}}{{100}}} \right)^2}\\\\ = {73.5^2}\\\end{array}$

$\begin{array}{l}\\ = 5402.25\\\\ \approx 5402\\\end{array}$

(2.d)

The margin of error of $100 has provided a sample size of 5402 in the previous part. But this sample size is not recommended as it is too large. Hence, do not recommend to obtain the $100 margin of error.

Ans: Part 1

The planning value for population standard deviation is $3,750.

Part 2.a

The sample size that should be taken for the margin of error of $500 is 216.

Part 2.b

The sample size that should be taken for the margin of error of $200 is 1351.

Part 2.c

The sample size that should be taken for the margin of error of $100 is 5402.

Part 2.d

Do not recommend to obtain the $100 margin of error.

Answer 2

Answer 3

solution in At asy. c. I, z=1.96 planning value = 45000-30000 15000 = 3750 b) = (1.96 x 3750 gr - (14.7) = 216.09 2216 = (1.96x of = (1.96x 2750 ) = (1350.56) ~ 1351 = 5402.25 Y suo2 d) from pard (C), we know that in order to satisfy the margin of requirement of $100 the sample size is required 5402, which is unmanageably large. So it is not recommended to obtain $100 margin of error.