The margin of error formula is
Assume that I created a 95% confidence interval for the mean hours studied for a test based on a sample of 100 students. The lower bound of this interval was 6 and the upper bound was 10. Assume that when I created this interval I knew the population standard deviation.
Calculate and report the population standard deviation.
I knew I use the E = zα/2 * (2√ n) formula (margin of error) but I know how to apply it to this problem. I also found that the sample mean is 8. Im just stuck on how to find the population standard deviation.
For a 95% confidence interval 47.5% of the population is below the mean and 47.5% is above the mean.
You can use this chart to figure out the Z value for the 95% interval Standard Normal Distribution Table
Oh! I see, that why I kept getting the wrong answer. So then I would just plug this number into the 'Z' portion of the equation then?
So it would look like this:
4= 1.96 (α/10) ?
Edit: My book is saying that the margin of error equation doesn't include the Z1-α/2 but only the Zα/2 part without 1-. Is that correct..?
The margin of error E is the difference between the sample mean and the upper limit, or the sample mean and the lower limit.
For a 95% confidence interval the significance level is 0.05 (the amount of the population not in the confidence interval)
. is the point in the normal distribution that has 2.5% of the population below it, that point is -1.96
. is the point with 97.5% of the population below it, that point is 1.96.
So it doesn't really matter, all that changes is the sign