Experimental calculation of solubility product constant In order to calculate the $K_\mathrm{sp}$ of lead(II) iodide, I mixed $\pu{6.0 mL}$ of $\pu{0.01 M}$ $\ce{Pb(NO3)2}$, $\pu{6.0 mL}$ of $\pu{0.02 M}$ $\ce{KI}$, and $\pu{8.0 mL}$ of water, and a precipitate (barely) formed.

From this, I calculated $[\ce{Pb^2+}]$ to be $\pu{0.0036 M}$ and $[\ce{I-}]$ to be $\pu{0.0060 M}$, giving

$$K_\mathrm{sp} = [\ce{Pb^2+}][\ce{I-}]^2 = 1.1 \times 10^{-7}.$$

However, the actual value (at $\pu{25 ^\circ C}$) is $7.9\times 10^{-9}$ (Skoog et al. Fundamentals of Analytical Chemistry), so my result is off by a factor of $\sim 10$.

I'm pretty sure I had clean equipment and measured carefully and accurately. Why is there a discrepancy? Is my maths wrong, or are there some inaccuracies in my experiment?
It's probably an experimental error. The 'barely' may be important here.

For one, you may not be sure how accurate your initial molarities are. A mistake in one of them ruins it all. Titrate them properly and check. Titration usually gives two significant digits.

Also, since its a cube overall, a small mistake can get much larger. You may want more accuracy in your readings (one significant digit isn't good enough), and you need to improve on the 'barely'. Though a factor of 11 needs a variation by a factor of 2.3 in both readings. So I don't know.

Have you redone the experiment? Doing an experiment more than once is integral to getting good values.