Question

Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $30,000 and $45,000 with a population standard deviation of $3750. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired. How large a sample should be taken if the desired margin of error is $200?

Answer 1

The value is
`n =1351`

Explanation:

From the question we are told that

The standard deviation is
`\sigma = \$3750`

The margin of error is
`E = \$ 200`

From the question we are told the confidence level is 95% , hence the level of significance is

`\alpha = (100 - 95 ) \%`

=>
`\alpha = 0.05`

Generally from the normal distribution table the critical value of
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  1.96`

Generally the sample size is mathematically represented as

`n = [\frac{Z_{(\alpha )/(2) } *  \sigma }{E} ] ^2`

=>
`n = [1.96  *  3750 }{200} ] ^2`

=>
`n =1351`