Complete Question
What is the minimum sample size required to estimate a population mean with 90% confidence when the desired margin of error is 1.25? The standard deviation in a pre-selected sample is 7.5
Answer:
The minimum sample size is [tex]n =97[/tex]
Step-by-step explanation:
From the question we are told that
The margin of error is [tex]E = 1.25[/tex]
The standard deviation is [tex]s = 7.5[/tex]
Given that the confidence level is 90% then the level of significance is mathematically represented as
[tex]\alpha = 100 - 90[/tex]
[tex]\alpha =10\%[/tex]
[tex]\alpha =0.10[/tex]
Next we obtain the critical value of [tex]\frac{\alpha }{2}[/tex] from the normal distribution table
The value is [tex]Z_{\frac{ \alpha }{2} } = 1.645[/tex]
The minimum sample size is mathematically evaluated as
[tex]n = \frac{Z_{\frac{\alpha }{2} * s^2 }}{E^2 }[/tex]
=> [tex]n = \frac{1.645^2 * 7.5^2 }{1.25^2 }[/tex]
=> [tex]n =97[/tex]
