In a sample of $15$ normally distributed random variables with unknown expectation $µ$ and variance $σ^2$, the sample mean was $10.3$ and the sample variance $s^2$ was $0.13.$ Make a symmetric confidence interval for $µ$ at the $99$% confidence level.
Well, the confidence interval is given by
$$\mu=\overline{X}\pm z\frac{s}{\sqrt{n}},$$
where $\Phi(z)=(1+q)/2.$ I have that
$$\Phi(z)=\frac{1+0.99}{2}=0.995 \implies z=\Phi^{-1}(0.995)=??$$
Which one of the ones I've marked should I choose?
The book says $z=2.977,$ which doesn't seem close to what I had in mind. Why this deviation?
The discrepancy is related to the $t$-distribution. Recall that the cost of replacing an unknown $\sigma^2$ with its estimate $S^2$ is that the resulting quantity $$\frac{\overline x - \mu}{S / \sqrt n}$$ has a student's $t$-distribution, not a normal distribution. (If you replaced $S$ with $\sigma$, then the quantity would have a normal distribution; also, as $n \to \infty$, the student's $t$-distribution approximates a standard normal distribution.)
As a result, you are consulting the wrong table. You instead need a table of values for a $t$-distribution with 14 df. You can verify that the book's calculator is correct with, for instance, this calculator (df = 14, $P(T \leq t) = 0.995$).