Finding Sample Standard Deviation Help

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.**

Re: Finding Sample Standard Deviation Help

The margin of error formula is

Re: Finding Sample Standard Deviation Help

Quote:

Originally Posted by

**Shakarri** The margin of error formula is

Yes, I know this but I don't know how to find the population standard deviation from the formula.

Re: Finding Sample Standard Deviation Help

Ok, you misstyped it in your first post.

Try using the formula for the upper bound of the confidence interval to get

Re: Finding Sample Standard Deviation Help

Quote:

Originally Posted by

**Shakarri** Ok, you misstyped it in your first post.

Try using the formula for the upper bound of the confidence interval to get

Oh, whoops! Haha. I tried but I get stuck on how to figure out the Z portion.

Re: Finding Sample Standard Deviation Help

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

Re: Finding Sample Standard Deviation Help

Quote:

Originally Posted by

**Shakarri** 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

So would the z value be 1.645?

Re: Finding Sample Standard Deviation Help

No, you were looking at the situation where 50% is below the mean and 45% is above the mean. When you have 47.5% above the mean Z is equal to 1.96

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Re: Finding Sample Standard Deviation Help

Quote:

Originally Posted by

**Shakarri** No, you were looking at the situation where 50% is below the mean and 45% is above the mean. When you have 47.5% above the mean Z is equal to 1.96

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..?

Attachment 30404

Re: Finding Sample Standard Deviation Help

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

Re: Finding Sample Standard Deviation Help

Quote:

Originally Posted by

**Shakarri** 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

So then would the formula I posted in my previous post be correct to solve for the population standard deviation?

Re: Finding Sample Standard Deviation Help

I keep getting an answer of 20.408 and I know that isn't correct..

Re: Finding Sample Standard Deviation Help

E is the difference between the sample mean and the upper limit, you were using the difference between the lower limit and the upper limit