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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.

  1. 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?
  2. 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.

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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.

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  • $\begingroup$ That's exactly what I did and I just realised my mistake, was taking the square root at the end instead of squaring both sides because I forgot you don't square $n$ for the variance. Thanks. $\endgroup$ – John Miller Mar 25 '19 at 14:27
  • $\begingroup$ @JohnMiller You´re welcome. Nice that it work now. Don´t forget to mark the answer as accepted. $\endgroup$ – callculus Mar 25 '19 at 15:21

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