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The context of the question is that a bakery bakes cakes and the mass of cake is demoted by $X$ such that $X sim N(300, 40^2)$. A sample of 12 cakes is taken and the mean of the sample is 292g. The question wants me to find the $p$-value and test to see if the mean has changed at 10% significance.

So I know how to carry out the test as $overlineX_12 sim N(300,frac40^212)$, But what would the p-value I'm trying to calculate be? I know the p-value is 0.244.

The hypothesis testing:
$$H_0: mu =300\
H_1:mu ne 300 \
z=fracbarx-30040/sqrt12=-0.6928\
ptext-value=P(z<-0.6928)=0.244 \
textReject $H_0$ if $p<fracalpha2$: 0.244not < 0.05 Rightarrow textFail to Reject H_0.$$
Note: $p$-value calculation:

1) In MS Excel: $=NORM.S.DIST(-0.6928;1)$.

2) WolframAlpha.

3) Z table.

$newcommandPmathbbP$It appears that you have $Xsim N(mu, 40^2)$ (known variance), and your null hypothesis is that $mu = 300$, with alternative hypothesis $mune 300$. For this, we will be using a two-tailed test.

Under the null hypothesis, we have $overlineX_12sim Nleft(300, frac40^212right)$, or $T := fracoverlineX_12 - 30040/sqrt12sim N(0,1)$. The $p$-value is the probability of getting a "more extreme" result for the test-statistic $T$ than observed under the null hypothesis (note the observed value is $colorbluefrac292-30040/sqrt12$), which for our two-tailed test means that the $p$-value is

$$Pleft(|T| > left| frac292-30040/sqrt12right| right) quad textwhere T sim N(0,1).$$

If you compute this probability, it seems you get double the answer you wrote. So the answers probably used a one-tailed test, which would be the case if our alternative hypothesis was $mu < 300$, in which case the $p$-value would be

$$Pleft(T < frac292-30040/sqrt12 right) quad textwhere T sim N(0,1).$$

This will get you the reported answer. The reason I used a two-tailed test is because I interpreted the alternative hypothesis as being with a $ne$ sign rather than $<$, because the wording of the question was "mean has changed".