Determining Yes or No?

Octave, 29 27 bytes

Thanks to @RickHithcock for pointing out a mistake, now corrected. Also, 2 bytes off thanks to @StewieGriffin!

@(s)'yn'(mod(sum(s+1),2)+1)

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Explanation

The ASCII code point of 'y' is odd, and that of 'n' is even. The code

1. adds 1 to each char in the input string to make 'y' even and 'n' odd;
2. computes the sum;
3. reduces the result to 1 if even, 2 if odd;
4. indexes (1-based) into the string 'yn'.

JavaScript (ES6), 28 bytes

Takes input as a string.

s=>'ny'[s.split`n`.length&1]

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JavaScript (ES6), 30 bytes

Takes input as an array of characters.

y=>'yn'[n=1,~~eval(y.join`^`)]

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Charcoal, 6 bytes

§yn№Ｓn

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    Ｓ   Input string
   № n  Count number of `n`s
§yn     Circularly index into string `yn`
        Implicitly print appropriate character