Interval Estimation questions I am completely lost at how to do this. My instructor talks so fast i can barely understand her so figuring this out was struggling. This website is my last choice and i hope others can help me out so i can understand.
I need to find the sample mean and the margin of error of a question, but im not sure if i can post the exact question because i got banned from stack overflow for doing that so i'll put it in another way ^^:
"Suppose that based on a random sample of 35 graduate students, a 95% confidence interval for the mean hours spent studying (per day) was found to be (3.5,3.9)"
a) find the margin of error
b)find the sample mean
for the margin of error i subtracted the two means to find the difference of .4 and divided by 2 to get the margin of error to be .2 (pretty sure thats wrong lol)
If anyone can help me out, i would be very thankful. also, please let me know if this type of question isnt allowed lol
 A: If you have a random sample $X_1, X_2, \dots, X_n$ from a normal population
and want a 95% confidence interval for the population mean $\mu,$ then a 95%
confidence interval is of the form $$\bar X \pm t^*\frac{S}{\sqrt{n}},$$ where
$\bar X$ and $S$ are the sample mean and standard deviation, respectively,
and $t^*$ cuts 2.5% of the probability from the upper tail of Student's t
distribution with $n - 1$ degrees of freedom. Alternatively, the confidence
interval is $(\bar X - t^*\frac{S}{\sqrt{n}},\, \bar X + t^*\frac{S}{\sqrt{n}}).$
The quantity $t^*\frac{S}{\sqrt{n}}$ is called the 'margin of error' of the confidence interval. In your Question the confidence interval is $(3.5,\,3.9),$
so the margin of error is $(3.9 - 3.5)/2 = 0.2$ as you say. Also,
$\bar X = (3.9 + 3.5)/2 = 3.7.$
Notes: 
(1) You say $n = 35$ in your problem so $t^* = 2.032$ in this case. This value from R statistical software as shown below, but
you should try to find the same value from a printed table of
t distributions in your book. (It's on row 'df=34' of the table; what does the corresponding column header mean?)
qt(.975, 34)
[1] 2.032245

(2) Here is a sample of size $n = 35$ from another normal distribution, along
with (slightly edited) output from R giving a 95% CI for the population mean
$\mu.$ You should be able to verify the stated CI $(95.14, 106.11)$ from other information in
the output. 
set.seed (1112);  x = rnorm(35, 100, 15)
summary(x);  sd(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  62.26   93.22  100.85  100.62  110.42  128.23 
[1] 15.96661  # This is the sample SD

t.test(x)     # Does a t test (edited out ...) and makes a 95% confidence interval

        One Sample t-test
...
95 percent confidence interval:
 95.13601 106.10546
sample estimates:
mean of x 
 100.6207 

(3) Be careful not to confuse the terminology: the quantity
$\frac{S}{\sqrt{n}},$ without the $t^*$-factor, is called the '(estimated) standard error of the mean'.
