Modelling number of boxes delivered in terms of number stated to be delivered I am looking for ways how to approach this problem. 
let say we have person A that tells me how many boxes I will receive this week. but this person A cannot be trusted, he always lies and maybe miscount his number.
but I can say for sure his number is always higher than the number of actual boxes I receive each week. If I keep a tally of what he said and the actual number, given enough time, wouldn't that mean I will be able to guess better at what the number is going to be? 
+---------+----------+--------+-------  ---+--------   ----------+
|  Week   | Person A | Actual | discrepancy| discrepancy percent |
+---------+----------+--------+----------  +-------------   -----+
| week 1  |       11 |      5 |        6   |      45.45454545    |
| week 2  |        9 |      9 |        0   |              100    |
| week 3  |       11 |      5 |        6   |      45.45454545    |
| week 4  |       11 |      7 |        4   |      63.63636364    |
| week 5  |        8 |      8 |        0   |              100    |
| week 6  |       11 |      9 |        2   |      81.81818182    |
| week 7  |       10 |      5 |        5   |               50    |
| week 8  |       10 |      7 |        3   |               70    |
| week 9  |       10 |      9 |        1   |               90    |
| week 10 |        8 |      7 |        1   |             87.5    |
| week 11 |        9 |        |            |                     |
+---------+----------+--------+----------  +-----------------   -+

let say this coming week 11, person A says, I will receive 9 boxes. how can I guesstimate the actual number based on the data I have in the past?
 A: Number of boxed delivered will be a count.  It is both integer and non-negative. You should consider that when looking at potential models.
I would expect the discrepancy may change in size with the person's stated numbers but I wouldn't expect that it would necessarily be linear (I'd anticipate that discrepancy would increase but do so more slowly than linear). This means that percentage discrepancy may not be the best way to model the relationship.
That is, I expect that (y-x)/x would not be constant, it would be a function of x -- but probably not a linear one.
Given these considerations, I'd be inclined to use a poisson glm or perhaps a negative binomial model to predict actual counts from the stated number to be delivered. I'd perhaps use a log-link and fit against log(x) to get a power-relationship (which would allow for a slower than linear increase). However, a linear fit would probably not do so badly either (still using Poisson or negative binomial likelihood though).
Perhaps something like this in R (I'm assuming the data has been read in as in nex's answer but into a data frame called boxes):
modelp <- glm(Actual~log(PersonA),data=boxes,family=poisson(link=log))
deliv.pred <- predict(modelp,newdata=data.frame(PersonA=4:11),type="response")
plot(Actual~PersonA,boxes,pch=16)
points(deliv.pred~I(4:11),col=4,type="o")
abline(0,1,col="darkred",lty=3)  # dot in the y=x line

With these sample data the coefficient was a little greater than $1$ so the fitted trend increases more quickly than linearly (though a pure linear fit would do about as well on those numbers). 

[From the fitted model on these data, the fitted discrepancy as a percentage of stated is actually decreasing -- the curve is trending away from y=x less strongly at higher "PersonA" values. There's not enough data to say whether this effect is really there or just an artifact of the model.]
You can see that the blue curve is a lot closer to the data than the reddish-brown dashed line, so the model has definitely improved the prediction over what PersonA gives.
I don't think this is likely to be an ideal model, but might be adequate. If you have multiple models you wish to look at you might consider looking at them on your oldest data and then fitting a chosen model to more recent numbers.
However, I'd also anticipate the possibility of a negative relationship to previous discrepancies -- if the stated number was farther than typical from what was delivered it might be closer next time and if it was closer, the next time you'd expect it to be further -- a kind of ethicality fatigue, so I'd consider looking for some kind of autocorrelation in (some kind of) residuals from this sort of model.
A: Harvey is right that you should rename your "variance" column to "discrepancy".  Assuming you accummulate enough values for your "discrepancy" variable, you will end up with a time series of "discrepancy" values, which would then put you in a forecasting context. 
Specifically, you could us the "historical" values of your "discrepancy" time series to forecast the next value of the time series.  This forecast exercise would give you the flexibility to consider various models to capture the temporal dynamics of the historical time series (e.g., ARIMA models, exponential smoothing models).  However, the key is having enough data to perform a meaningful forecast.  If your historical time series is quite short, you'll have to consider much simpler models than if it's quite long.  How well your forecast will work will ultimately depend on the level of "noise" in your data - you may be able to reduce that "noise" by incorporating some relevant explanatory variables in your model. 
This time series framework will also enable you to compare your forecasted "discrepancies" against the actual "discrepancies" (once they become available) so that you can keep track of whether or not your forecasting ability improves/degrades over time.
The linear regression approach already suggested has the drawback that it doesn't account for the potential temporal correlation of the values of your Actual variable.  That unaccounted for correlation can "dilute" the signal you are after. 
A: You can build a linear regression model and then use it to predict Actual from PersonA.
For example in R:
# create dummy data.frame
mat=data.frame(PersonA=c(11,4,11,4,5,6),Actual=c(8,1,9,3,3,5))
# fit the model
fit=lm(Actual~PersonA,data=mat)
# new prediction
new.est=data.frame(PersonA=c(11))
# estimate based on the new prediction
predict(fit, new.est)

A: First of all, your labeling could be better. "Variance" has a specific meaning in statistics, so you should use another term, such as "discrepancy". Your "variance percent" is actual as percent of asserted. If you want discrepancy as percent of asserted, you would need to take 100% minus those numbers. This is just a labeling issue; the p-value calculations aren't affected.
Modeling is all about making assumptions, and with so little data, one's assumptions are going to dominate the results. The simplest assumption is to pick a feature and assume that it has a constant mean. The correlation between asserted and actual is -.329. For 10 data points, there's a 33% probability of that happening by chance. The correlation between actual as percent of asserted versus asserted is .711. There's a 2% probability of that happening by chance. For asserted and discrepancy, the correlation is .74. So of those hypotheses, the one that fits the data the best is that the actual amount is independent of the asserted. The actual amount has a mean of 7.1 and a standard deviation of 1.577, so you should expect to receive about 7 each month, plus or minus around two, and knowing the asserted does not add a statistically significant amount of information. 
The high correlation between discrepancy and asserted might make it seem like you can predict the discrepancy from the asserted, but the reason there's a correlation is because the higher the asserted is, the higher the discrepancy is, for a constant actual. If you try to do a linear regression of the asserted and predict the discrepancy, your errors will be exactly the same as if you just predict actual based on asserted.
As you get more data, you'll be able to draw better conclusions. As it stands, it appears that as the asserted number increases, the actual number decreases, but this is consistent with random variation.
Also, the difference between the asserted an actual have a mean of 2.8 and a standard deviation of 2.22. Assuming a normal distribution with those parameters, the probability of the difference being positive is thus 89.6% for a particular value. The probability of all ten being positive is 33%. So the observed phenomenon of the asserted being larger than the actual is consistent with the hypothesis that this is due simply to random chance. That is, it may just have been random chance that the asserted were previously larger than actual, and actual larger than asserted are possible.
