Margin of error and average for a sample. In a sample of Petri dishes the number of possible infectious microorganisms was counted, obtaining the following results after 11 counts.
3,6,7,2,4,7,8,9,10,2,5
I have to prove that.
I) The average number of microorganisms is significantly equal to 4.
Well,the average of this sample is equal to 5.72, but how do I find the significance?
II) The margin of error of the confidence interval is greater than 2.
I haven't worked with margin of error before but I searched and it says that for a CI it's the length divided by 2
For an $CI=[5.72-1.96\sqrt(6.926),5.72+1.96\sqrt(6.926)] $, so the length is 10.25 and divided by two the margin error is 5.12.
Is this right?
UPDATE:
For the part I) I was thinking about the CI, but this time using the t distributions because of the size of the sample. So if 4 belongs to the CI is significative?
 A: By doing a test, you cannot "prove" anything in a strictly mathematical sense.
Also, you seem to be doing one-sample t procedures [or possibly z procedures], and there seems to be
no reason to believe that the data are from a normal population. 
Nevertheless, t procedures are not extremely sensitive to non-normality unless the sample is severely skewed or shows far outliers, which your sample does not. [It would be incorrect to use z procedures here because the population standard deviation $\sigma$ is unknown.]
Here is a boxplot of your eleven observations:

Below are results from a one-sample t test in R statistical software, which also gives a 95% confidence interval for the population mean $\mu$
(along with a few related computations).
Test of hypothesis: The null hypothesis is $H_0: \mu = 4$ and the alternative hypothesis is $H_a: \mu \ne 4.$ The P-value (0.065) exceeds 0.05, so results are "consistent with" $H_0$ at the 5% level of significance. In other words $H_0$ cannot be rejected at the 5% level.
Confidence interval: This does not imply that $\mu = 4,$ only that we have no evidence that
$\mu \ne 4.$ Indeed, the 95% confidence interval $(3.87, 7.58)$ indicates that values $4.0 \pm 3.58$ should be considered as realistic
values of $\mu.$ The margin of error $M = 3.58$ does indeed exceed 2.
You are correct that the sample mean is $\bar X = 5.72,$ but the sample standard deviation is $S = 2.76,$ which does not seem to match what you have. The correct method of computing the CI is $\bar X \pm t^*S/\sqrt{11},$ where $t^* = 2.228$ cuts 2.5% of the probability from the upper
tail of Student's t distribution with $n - 1 = 11 - 1 = 10$ degrees of
freedom. 
Output from R statistical software:
t.test(x, mu=4)

        One Sample t-test

data:  x
t = 2.0755, df = 10, p-value = 0.06468
alternative hypothesis: true mean is not equal to 4
95 percent confidence interval:
 3.873009 7.581537
sample estimates:
mean of x 
 5.727273 

sd(x)
## 2.760105
qt(.975, 10)
## 2.228139
mean(x) + pm*qt(.975, 10)*sd(x)/sqrt(11)
## 3.873009 7.581537  # matches CI in output from t.test

Note: (1) Below is a printout from Minitab statistical software
showing the t test and confidence interval. This output agrees with the output from R.
One-Sample T: x 

Test of μ = 4 vs ≠ 4

Variable   N   Mean  StDev  SE Mean      95% CI         T      P
x         11  5.727  2.760    0.832  (3.873, 7.582)  2.08  0.065

(2) In case you are concerned about using a t confidence interval for data not known to be normal, a 95% Wilcoxon confidence interval
for the population median $\eta$ is $(3.50, 8.00),$ which exceeds 4 units in length. This procedure assumes symmetry of the population, but not normality. Also, a 95% nonparametric bootstrap confidence interval for $\mu$ is $(4.18, 7.27),$ which makes no assumption of normality. However, bootstrap procedures on samples as small as $n = 11$ are of dubious accuracy.
