I was trying to solve the following exercise:
"Given a population $\Omega$, normally distributed, and its sample $C=\left\{5,10,6,14\right\}$, calculate the confidence interval of the population mean with a $90\%$ significance level. Afterwards, test with a $1\%$ significance level if the population mean might be supposed equal to 10.$
First of all, I calculated the sample mean $$\bar{X}=35/4,$$ then the unbiased sample variance $$S^2=\frac{203}{16},$$ and finally I wrote the confidence interval $$\left|\frac{\bar{X}-\mu}{\frac{S}{\sqrt{n}}}\right|<t_{0.90}(3)\Rightarrow-\frac{\sqrt{n}}{S}t_{0.90}+\bar{X}<\mu<\frac{\sqrt{n}}{S}t_{0.90}+\bar{X}.$$ Is that correct? I'm wondering if the "smallness" of the sample requires an altnernative procedure to solve the esercise. Then I completed the latter calculating the statistics $$T=\frac{\bar{X}-\mu_0}{\frac{S}{\sqrt{n}}}\simeq -0.68$$ and I stated the the null hypotesis should not be rejected since $$|-0.68|<t_{0.975}(3)=3.182.$$ Any suggestion would be highly appreciated. Thanks in advance!
The smallness of the sample is precisely the reason why it's important to use the t-distribution rather than the normal distribution, in this case with $3$ degrees of freedom.
$\require{cancel}$ \begin{align} \text{wrong: } & \xcancel{-\frac{\sqrt n}S t_{0.95}+\bar{X}<\mu<\frac{\sqrt n}S t_{0.95}+\bar{X}. \vphantom{\frac{\displaystyle\int}{\displaystyle\int}} } \\[10pt] \text{right: } & -\frac S {\sqrt n} t_{0.95}+\bar{X}<\mu< \frac S {\sqrt n} t_{0.95}+\bar{X}. \end{align}