Question

Using the same sample data, the margin of error for an 80% confidence interval is larger than the margin of error for a 90% confidence interval. True or false? why?

I thought it was false, but I think I might be wrong.

Answer 1

Answer

Margin of error = z*(SD/sqrt(n))

z is critical z score value, SD is standard deviation and n is the sample size

it is given that the sample data used is same, i.e. SD and n are same for 80% and 90% confidence levels

z critical value for 80% confidence interval and 90% confidence interval using z distribution table are 1.28 and 1.645 respectively

So, Margin of error for 80% = 1.28*(SD/sqrt(n))

and Margin of error for 90% = 1.645*(SD/sqrt(n))

We know that 1.645 is larger than 1.28, So the margin of error for 90% confidence interval will be larger than 80% confidence interval

Therefore, given statement is false