in= [3 3; 3 3]
in =
3 3
3 3
>> sum(in.*isprime(in))/sum(isprime(in))
ans =
3.0000
The fault is in two respects:
(1) the numerator and denominator for Test 3 will be vectors, not scalars, because that's how sum() works;
(2) the MATLAB algorithm for matrix division is therefore being employed, and it has introduced a truncation error. Your code yields a purported answer of 3 – (4.4409E–16) for Test 3. Of course, if only a few decimal places are displayed, this is shown as "3.0000".
To avoid this problem, reshape the input matrix to a vector. This can be done with the reshape command, but an easier way is to simply index as "in(:)".
This is done in https://au.mathworks.com/matlabcentral/cody/problems/14findthenumericmeanoftheprimenumbersinamatrix/solutions/1142672
Another effective (but somewhat less elegant) way of avoiding the problem is to nest your summations: sum(sum(sum( ... ))). (But you must use at least as many "sum" command as your matrices have dimensions.)
For debugging, try using the whos command.
Test  Status  Code Input and Output 

1  Pass 
x = 3;
y_correct = 3;
assert(isequal(meanOfPrimes(x),y_correct))

2  Pass 
x = [1 2 3];
y_correct = 2.5;
assert(isequal(meanOfPrimes(x),y_correct))

3  Fail 
x = [3 3; 3 3];
y_correct = 3;
assert(isequal(meanOfPrimes(x),y_correct))

4  Pass 
x = [7 3 8 8]';
y_correct = 5;
assert(isequal(meanOfPrimes(x),y_correct))

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Project Euler: Problem 6, Natural numbers, squares and sums.
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