Answer:
The margin of error is 3.465
Step-by-step explanation:
Margin of error = t × sd/√n
n = 92, degree of freedom = n - 1 = 92 - 1 = 91, t-value corresponding to 91 degrees of freedom and 90% confidence level is 1.6618, sd = 20
Margin of error = 1.6618×20/√92 = 33.236/9.592 = 3.465
Answer:
The margin of error is of 3.43.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.9}{2} = 0.05[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.05 = 0.95[/tex], so [tex]z = 1.645[/tex]
Now, find the margin of error M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
So
[tex]M = 1.645*\frac{20}{\sqrt{92}} = 3.43[/tex]
The margin of error is of 3.43.
