Answer:
The margin of error is [tex]ME = 2.2962[/tex]
Step-by-step explanation:
From the question we are told that
The number of student is [tex]n = 500[/tex]
The highest amount is [tex]A =[/tex]$200
The lowest amount is [tex]B =[/tex] $75
The sample mean is [tex]x =[/tex] &140
The Standard deviation of this set is mathematically evaluated as
[tex]s = \frac{A-B}{4}[/tex]
Substituting values
[tex]s = \frac{200-75}{4}[/tex]
[tex]s = 31.25[/tex]
The margin of error (ME) is mathematically evaluated as
[tex]ME = t_{n-1}__{\alpha }} * \frac{s}{\sqrt{n} }[/tex]
Where [tex]t_{n-1}__{\alpha }}[/tex] is the critical value for [tex]\alpha = 0.05[/tex] i.e the significance level
From the critical value table this is [tex]t_{n-1}__{\alpha }} = 1.649[/tex]
So
[tex]ME = 1.649 * \frac{31.25}{\sqrt{500} }[/tex]
[tex]ME = 2.2962[/tex]
