How to calculate P-value for location normal? Marks on an exam in a statistics course are assumed to be normally distributed with unknown mean but with variance equal to $5$. A sample of four students is selected,and their marks are $52, 63, 64, 84$. 
Suppose that we drop the assumption that the population variance is $5$.  Assess the hypothesis $H_0$: $u=60$ by computing the relevant P-value and compute a $0.95$-confidence interval for the unknown $u$

I'm just looking to assess it.
I'm not sure what the formula is for this but this is what I did according to the one my textbook used:
$\bar{x} = 65.75$ (average)
$u = 60$ (given)
$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} = 13.3$
So the $P$-value formula is given by
$$P\text{-value} = P\left(t_{df=3} \leq \left| \frac{\bar{x}-u}{s/\sqrt{n}} \right|\right) = P\left(t_{df=3} \leq \frac{65.75-60}{13.3/2} \right) = 2 \left(1-\Phi(0.864) \right) = 2(1-0.75) = 0.5 > 0.05$$ so we accept. 
But this is completely wrong according to solution they got $t-statistic$ equals 9.12 ?? and $p-value$ equals 0.452 ?? 
 A: Yes, when you have a sample size which is small and the population standard deviation is known, the Student's $t$ distribution estimates the mean if you assume the population is normally distributed.
You have 4 points, so you use the $t$ distribution test with $4-1=3$ degrees of freedum.  So you take the sample mean minus hypothesis mean of $65.75-60.00 = 5.75$ divided by the sample standard deviation $13.3$ and multiplied by $\sqrt{n} = 2$.
This gives $0.8647$.  The Cumulative Distribution function for $0.8646$ with $\nu = 3$ is 
$0.782$. That means that there is about a 21.8% chance that the deviation from $60$ would be worse than the observed deviation, so you would accept (or more correctly, fail to reject) the hypothesis that the actual mean is $60$. 
A: Starting with $X\sim N(?,\sigma^2)$, you can get $Y=X/\sigma, Y\sim N(?,1)$. Then, we want to test $H_0:?=0$, so we want to find how surprised we should be to get the sample we got using that ($H_0$) assumption. If we were to observe $y=1.965$ in this single sample example, we find $P(|Y|\ge 1.96)<0.05$ and we can reject the $H_0$ in a 2-sided test.
In the multiple-sample setting, you need to find a similar "pivot" ($Y$ in the example I wrote above) for which you know the distribution. Then, find $P(theoretical pivot \ge observed pivot)$.
