Hypothesis testing for sample means within a normal distrubution

Question

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 $\overline{X_{12}} \sim N(300,\frac{40^2}{12})$, But what would the p-value I'm trying to calculate be? I know the p-value is 0.244.

Answer 1

$\newcommand{\P}{\mathbb{P}}$It appears that you have $X\sim N(\mu, 40^2)$ (known variance), and your null hypothesis is that $\mu = 300$, with alternative hypothesis $\mu\ne 300$. For this, we will be using a two-tailed test.

Under the null hypothesis, we have $\overline{X}_{12}\sim N\left(300, \frac{40^2}{12}\right)$, or $T := \frac{\overline{X}_{12} - 300}{40/\sqrt{12}}\sim N(0,1)$. 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 $\color{blue}{\frac{292-300}{40/\sqrt{12}}}$), which for our two-tailed test means that the $p$-value is

$$\P\left(|T| > \left| \frac{292-300}{40/\sqrt{12}}\right| \right) \quad \text{where }T \sim N(0,1).$$

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

$$\P\left(T < \frac{292-300}{40/\sqrt{12}} \right) \quad \text{where }T \sim N(0,1).$$

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".

Answer 2

The hypothesis testing: $$H_0: \mu =300\\ H_1:\mu \ne 300 \\ z=\frac{\bar{x}-300}{40/\sqrt{12}}=-0.6928\\ p\text{-value}=P(z<-0.6928)=0.244 \ \\ \text{Reject $H_0$ if $p<\frac{\alpha}{2}$}: \ 0.244\not < 0.05 \Rightarrow \text{Fail to Reject} \ H_0.$$ Note: $p$-value calculation:

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

2) WolframAlpha.

3) Z table.