I was solving this question:
The mean IQ of a sample of 1600 children was 99. Is it likely that this was a sample from a population with mean IQ 100 and standard deviation 15?
like this:
Here level of significance (α) is not given, so I took 5% level.
Null hypothesis (H0): sample has been drawn from the population with mean IQ 100
Alternative hypothesis (H1): sample has not been drawn from the population hypothesis
Calculation:
n = 1600, $\bar{x}$ = 99 , $\mu$ = 100 , $\alpha$ = 15
$Z = \frac{\bar{x} - \mu}{\sigma_\bar{}/sqrt(n)}$ = 2.67
Cleary z > 1.96 , so H0 is rejected.
But since 2.67 gives 99.6% confidence level shouldn't we say that it is likely that the sample has been drawn from this sample ?
Although, the answers says it is not drawn from the sample.
As requested in comments:
This is confusing description, as a 99.6% confidence level does not means what it appears to mean compared to a 95% confidence level.
In other words, it makes you more confident that null hypothesis will not be rejected when it is correct, but less confident that it will be rejected when it is false.
You chose α=5% and that led you to reject the null hypothesis; that seems to be what the question wants