Question

We wish to construct a 90% confidence interval on the mean of a population known to be normally distributed based on a sample of 20 observations having a sample mean of 8 and a sample standard deviation of 2.
What is the margin of error for this confidence interval?

Answer 1

Answer:

Margin of error = 0.773

Step-by-step explanation:

We are given the following information in the question.

Sample size, n = 20

Sample mean = 8

Sample standard deviation = 2

90% Confidence interval

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$  

Putting the values, we get,  

$t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.10} = \pm 1.729$  

Margin of error:

$t_{critical}\displaystyle\frac{s}{\sqrt{n}}$  

Putting the values, we get,  

$1.729(\frac{2}{\sqrt{20}} ) = 0.773$