Amplitude reduction of Simple Moving Average on sinusoid
Below is the analytic result for both the actual max value of $0.901243$ and the maximum value found by the OP of $0.898464$

The reason you are not getting the predicted maximum is your samples of the sine wave are not located exactly at the peak. This is clear if you zoom in on the plot and compare the two peak locations for the number of samples given (as I show in the plot below).

 

Also a direct approach to establish the frequency response for the moving average filter that properly accounts for both the phase and magnitude of the filter is as follows:

Consider the implementation of a moving average filter given as ($1/N$ scaling not shown):



The general transfer function for the moving average filter directly from the implementation block diagram (where $z^{-1}$ is the transfer function of a unit delay) is:

$$H(z) = \frac{1}{N}\sum_{n=0}^{N-1}z^{-n}$$

Using the well known relationship for the geometric series (see at bottom of answer) this is:

$$H(z) = \frac{1}{N}\frac{1-z^{-N}}{1-z}$$

And the frequency response (as the DTFT of the impulse response) is a continuous function of frequency $\omega$ for all $z = e^{j\omega}$, therefore the values of z on the unit circle.  (Yes even though it is a discrete system the frequency response is indeed a continuous function and unique for $\omega$ over the range of $0$ to $\pi$ for real functions): 

$$H(\omega) = \frac{1}{10}\frac{1-e^{-j\omega N}}{1-e^{-j\omega}}$$

$$=\frac{1}{N}\frac{e^{-j\omega N/2}(e^{+j\omega N/2} - e^{-j\omega N/2})}{e^{-j\omega /2}(e^{+j\omega /2} - e^{-j\omega /2})}$$

Using Euler's identity relating $\sin$ to the positive and negative exponential terms (see at bottom of answer) results in:

$$ H(\omega)= \frac{e^{j((N-1)/2)}}{N}\frac{sin(\omega N /2)}{sin(\omega/2)}$$

The exponential term has a magnitude of 1 for all $\omega$ but provides for the exact phase shift between input and output as evidenced in the plots. 

The frequency $\omega$ is the normalized radian frequency in units of radians/sample, so in the OP's example $N=10$ and $\omega = \frac{2\pi}{40} = \frac{\pi}{20}$, and for$H(\pi/20)$ the result is:

$$H(\pi/20) = \frac{e^{j(4.5\pi/20)}}{10}\frac{sin(\pi/4)}{sin(\pi/40)} \approx 0.90124e^{-j0.70686}$$

Thus has a magnitude of $0.90124$ and an angle of $-.70686$ radians or -$40.500°$.

This result matches that provided by the freqz command in MATLAB/Octave:

&gt;&gt; h = freqz(ones(10,1),10, [0 pi/20]);
h = 1.0000 + 0.0000i  0.68351 - 0.58531i
&gt;&gt; abs(h(2))
ans = 0.90124
&gt;&gt; angle(h(2))
ans = -0.70686


Knowing the phase and amplitude from above, we can predict the result the OP got for the specific max sample point (15th sample) from $Asin(\omega n + \phi)$ using the amplitude $A$ and phase $\phi$ from above:

$$0.90124\sin(15 \pi/20 - 0.70686) = 0.898464$$

And the overall frequency response with magnitude in dB for $\omega = 0$ to $\pi$ is:



Note even though the samples of the output do not land on the exact peak does not mean this is not the peak of the output waveform. Consider Nyquists sampling theorem and how a waveform can be completely described (and if a single tone that would be its amplitude, frequency, phase etc) with relatively very few samples. 



Relationships Used Above

Geometric Series

$$\sum_{n=0}^{N-1}r^k = \frac{1-r^N}{1-r}$$

Euler's Identity for sine

$$sin(\theta) = \frac{e^{+j\theta}-e^{-j\theta}}{2j}$$