Test the claimed bulb power and change the sample size

Let $\mu$ be population mean power in watts. Treat 13 as the claimed null value to be tested: $H_0:\mu=13$, $H_1:\mu>13$. Under $H_0$ and independent normal sampling, $\bar X\sim N(13,2.716^2/50)$ exactly. Then $z=(13.168-13)/(2.716/\sqrt{50})=0.4373856378$. The upper-tail p-value is $1-\Phi(z)=0.3309158491\approx0.331>0.05$. Do not reject: there is insufficient evidence at 5% that the manufacturer understates the mean.

The 5% upper critical value is $\Phi^{-1}(0.95)=1.644853627\approx1.6449$. Reject for $z\ge1.644853627$; the observed 0.43739 lies outside the rejection tail. The source calculator uses Z-Test → Stats, $\mu_0=13$, $\sigma=2.716$, $\bar x=13.168$, $n=50$, alternative >, then Calculate. Its output $z\approx0.4373856378$, $p\approx0.3309158761$ has the same printed conclusion.

If the true population mean is 13 watts, applying this rule to repeated independent samples would wrongly conclude that the manufacturer understates the mean in 5% of cases. This is a conditional Type I error rate, not a 5% posterior probability that this conclusion or the null hypothesis is false.

At 10%, use $z_{0.90}=1.281551566$. For the same mean 13.168 and known SD 2.716, rejection needs $0.168\sqrt n/2.716\ge1.281551566$, so $n\ge429.2528020$ and the least integer is 430. The source rounds the critical value to 1.2816 and the bound to 429.3, with the same integer. A new larger sample is not guaranteed to reproduce the old mean; this is a threshold calculation conditional on that mean, not a power calculation.