Answer:
(507.05, 592.95)
Step-by-step explanation:
Given data:
sample mean = $550, sample standard deviation S = $60.05
95% confidence interval , n = 10
For 95% confidence interval for the mean
mean ± M.E.
where M.E. is margin of error = [tex]t_{n-1}, \alpha/2\times\frac{S}{\sqrt{n} }[/tex]
Substituting the values in above equation
[tex]=t_{10-1}, 0.05/2\times\frac{60.05}{\sqrt{10} }[/tex]
= 2.62×18.99
=42.955
= 550±42.95
=(507.05, 592.95)
