Is my reasoning on the following exercise correct? I am not sure if I correctly used Student t distribution instead of normal distribution and whether my calculation of std makes sense.
Exercise: "Four samples are taken and averaged every hour from a production line and the hourly measurement of a particular impurity is recorded. Approximately one out of six of these averages exceeds 1.5% when the mean value is approximately 1.4%. State assumptions that would enable you to determine the proportion of the individual readings exceeding 1.6%. Make the assumptions and do the calculations. Are these assumptions likely to be true? If not, how might they be violated?"
My attempted solution:
$n=4$, $df=3$
$\mu = 1.4\%$
$P(x>1.5\%) = 1/6$
$P(x>1.6\%) = ?$
Since the sample size is small and we don't know the population std, we need to use the Student-t distribution (with 3 degrees of freedom). For that, we need to assume that the samples are IID.
To compute the t-score we need $\sigma$, which is not given, but we can compute it from the information that $P(x>1.5\%) = 1/6 \approx 0.167$. It means that we can compute the value of t-statistic with inverse cdf of $1-0.167=0.833$. Inverse cdf for t-distribution with 3 degrees of freedom for $0.833$ is approximately $1.15$. From that we know:
$$1.15 = \frac{x-\mu}{\sigma/\sqrt{n}} = \frac{1.5\%-1.4\%}{\sigma/\sqrt{4}} = \frac{0.1\%}{\sigma/2}$$
hence $\sigma = 0.174\%$.
Now we can compute the t-score for $1.6\%$ which is:
$$t = \frac{1.6\%-1.4\%}{0.174\%/2} \approx 2.3$$
Next, reading from t-tables, we can check the value of $P(x>1.6\%)=P(t>2.3) \approx 0.05$, so there is around $5\%$ chance of getting a reading greater than $1.6\%$.
The assumption of IID could be not true if the samples were taken from different sources, or if the samples were not independent.
By assuming that
we have that the hourly average $\bar{X} \sim N(1.4,\frac{\sigma^2}{4})$. Then, from
$$P(\bar{X}>1.5\%) = P \left (Z=\frac{\bar{X}-1.4}{\sigma/\sqrt{4}}>\frac{1.5-1.4}{\sigma/\sqrt{4}} \right) =1-\Phi\left (\frac{1.5-1.4}{\sigma/\sqrt{4}} \right)=1/6,$$
you can obtain $\sigma^2$, and then you can compute
$$P(\bar{X}>1.6\%).$$
Working with T distribution is not needed here as we can directly compute $\sigma^2$.
If the sample is taken fully randomly at a certain time, in practice, only the first assumption may be violated, and $\bar{X}$ may not follow $N(1.4,\frac{\sigma^2}{4})$ as the sample size is only $4$. For a larger sample size ($\ge 30$), we can use the CLT. Note that for some population distributions (such as uniform distribution) even for $n=4$, the distribution of $\bar{X}$ becomes very similar to a normal distribution.