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

Margin of error = 0.773

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(s)/(√(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(s)/(√(n))`

Putting the values, we get,

`1.729((2)/(√(20)) ) = 0.773`