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To determine whether a prep course improved ACT scores, researchers collected a random sample of ACT scores from 1500 U.S. high school juniors who completed a prep course prior to taking the ACT exam. They also collected a random sample of ACT test scores from 1500 U.S. high school students who did not complete a prep course prior to taking the ACT exam. Those who completed the prep course had a mean score of 22.1 with a standard deviation of 3.9 and those who did not take the prep course had a mean score of 21.3 with a standard deviation of 4.3.

a. Determine the 95% confidence interval for both population means. b. Is the difference between the 2 population means statistically significant?

I used s.d. = $s/sqrt(n)$ to find the sample standard deviation, but this didn't seem to produce reasonable results. 0.1 for first group and 0.11 for second group. Lost as to the approach on this problem. Is this really math?

I understand that IF I could find the standard deviation I'd look at 2 std deviations to either side of the mean for the 95% interval. Not sure what to do next.

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    Does the difference lie in the confidence interval, if yes then that is expected and thus not statistically significant. If it doesn't then it is statistically significant to make you think otherwise on a confidence interval of $95$%. – 2017-02-02
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    Difference of what. I calculate a 95% confidence interval for both groups. What do I compare? – 2017-02-02
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    You need to look at the confidence interval of the difference between the means, have a look at this http://www.stat.yale.edu/Courses/1997-98/101/meancomp.htm – 2017-02-02
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    So I calculated the s.d of group 1 (with prep) using s.d $=s/sqrt(n)$ and got .1 and the second group I got 0.11 (not much difference. So, I then subtracted twice this value (to get to the 95% range) from each of the means. This gives me 2 ranges: 21.08 to 21.52 group 1 and for group 2: 21.9 to 22.3. Are these steps correct? what is next step? – 2017-02-02
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    Do you have a moment to chat in the forum about this? – 2017-02-02
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52912/discussion-between-user163862-and-ahmed-s-attaalla). – 2017-02-02
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    Please invite me @user163862 – 2017-02-02
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    The null hypotheses Is that there is no difference in mean. But from the confidence intervals no matter what two numbers we pick, we can never get a difference of means of $0$. So there is statistical significant evidence to suggest our null is false. – 2017-02-02
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    Btw calculating the standard deviation for the difference or sum of two things is easy. You can almost imagine it as using "Pythagoras's". If one thing has a standard deviation of $3$, and another $4$, then the standard deviation of the difference or sum of the two things is $\sqrt{3^2+4^2}=5$. This comes straight from $\text{var} (x+y)=\text{var }(x)+\text{var} (y)$ where var is variance, the square of standard deviation. – 2017-02-02

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