Question

The results of a survey has a confidence interval of 74.9% to 79.1%, 9 times out of 10. Determine the margin of error..

Answer 1

The lower limit of the confidence interval is:
\[\bar{x}-z. \sigma _{\bar{x}}=74.9\ .........(1)\]
The upper limit of the confidence interval is:
\[\bar{x}+z. \sigma _{\bar{x}}=79.1\ ..........(2)\]
where x-bar is the sample mean, sigma-xbar is the standard deviation of X-bar and z is 1.645 for a 90% confidence level.
The margin of error is given by the expression
\[z. \sigma _{\bar{x}}\]
Subtracting equation (1) form equation (2) gives:
\[2\times z .\sigma _{\bar{x}}=4.2 ......(3)\]
Now you can find the margin of error from equation (3)