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
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
Using the same sample data, the margin of error for an 80% confidence interval is larger...
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