The school canteen sells coffee in cups claiming to contain 250 ml. It is known that the amount of coffee in a cup is normally distributed with standard deviation 6 ml. Adam believes that on average the cups contain less coffee than claimed. He wishes to test his belief at 5% significance level.
- Adam measures the amount of coffee in 10 randomly chosen cups and finds the average to be 248 ml. Can he conclude that the average amount of coffee in a cup is less than 250 ml?
- Adam decides to collect a larger sample. He finds the average to be 248 ml again, but this time this is sufficient evidence to conclude at the 1% significance level that the average amount of coffee in a cup is less than 250 ml. What is the minimum sample size he must have used?
While 1. is pretty straightforward, I'm not sure what to do for 2. Since, $$ \bar{X} \sim N(250,\frac{6^2}{n}) $$ following the $z$-test, we can say $$ P(\bar{X} < 248) < 0.01 $$ to reject $H_0$, and then I was thinking of using, $$ z = \frac{x-\mu}{\sigma} $$ where $z = \Phi^{-1}\left(0.01\right)$ and solving for $n$ but this gives $n = 2.64$ which is clearly incorrect from 1. I don't necessarily want the answer, I would just like a bit of direction thank you.
You have $P(\bar{X} < 248) < 0.01$ which is
$\Phi\left(\frac{248-250}{\frac{6}{\sqrt{n}}} \right)<0.01$
$\Phi\left(\frac{-2\cdot \sqrt{n}}{6} \right)<0.01$
I think you can proceed.