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The scenario is that a group of $n$ people carry out this same experiment.

Each person generates 11 random numbers based on the normal distribution $\mu=2.7$ and $\sigma=0.6$.

Each person must then calculate their own test statistic for the hypothesis:

$$H_0:\mu=2.7$$$$H_1:\mu\neq2.7$$

When I carried out this test using Excel, my test statistic was $-3.19$ which is less than the critical value at a 5% significance level ($-1.96$). Therefore I would reject the null hypothesis. However, this is only my experiment, how would I then calculate the proportion of the group I would expect to reject $H_0$?

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The probability that one wrongly rejects the null hypothesis, when it is in fact correct, is the significance level, in this case $0.05$.

So if the experiment is repeated independently $n$ times, the expected number of times that the null hypothesis is wrongly rejected is $(0.05)n$.

The expected proportion of the group that wrongly rejects the null hypothesis is $0.05$.

Remark: You were so "unlucky" that it might be worthwhile to check the calculation that led to being $3.19$ standard deviation units away from the mean.

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