What you are dealing with in this question is a two-sided variance test, which is a specific case of a two-sided test with an asymmetric null distribution. The p-value is the total area under the null density for all values in the lower and upper tails of that density that are at least as "extreme" (i.e., at least as conducive to the alternative hypothesis) as the observed test statistic. Because this test has an asymmetric null distribution, we need to specify exactly what we mean by "extreme".
Lowest-density p-value calculation: The most sensible thing method of two-sided hypothesis testing is to interpret "more extreme" as meaning a lower value of the null density. This is the interpretation used in a standard likelihood-ratio (LR) test. Under this method , the p-value is the probability of falling in the "lowest density region", where the density cut-off is the density at the observed test statistic. With an asymmetric null distribution, this leads you to a p-value calculated with unequal tails.
To implement this method for the two-sided variance test, let $\text{Chi-Sq}(\cdot | n-1)$ denote the chi-squared density function with $n-1$ degrees-of-freedom. For a given test statistic $\chi_\text{obs}^2$, the p-value for the two-sided test is given by:
$$p(\chi_\text{obs}^2)
= \mathbb{P} \Big( \text{Chi-Sq}( \chi^2 | n-1) \leqslant \text{Chi-Sq}( \chi_\text{obs}^2|n-1) \Big| H_0 \Big).$$
In your particular problem you have $n=18$ and $\chi_\text{obs}^2 = 15.35667$. At this observed value of the test statistic you have density $\text{Chi-Sq}( \chi_\text{obs}^2|n-1) = \text{Chi-Sq}( 15.35667|17) = 0.07188203$, and this density cut-off also occurs at the lower point $\chi^2=14.64890$. This means that your two-sided p-value can be calculated as:
$$\begin{equation} \begin{aligned}
p(\chi_\text{obs}^2)
&= \mathbb{P} \Big( \text{Chi-Sq}( \chi^2 | 17) \leqslant 0.07188203 \Big| H_0 \Big) \\[6pt]
&= \mathbb{P} \Big( \chi^2 \leqslant 14.64890 \Big| H_0 \Big) + \mathbb{P} \Big( \chi^2 \geqslant 15.35667 \Big| H_0 \Big) \\[6pt]
&= 0.3792454 + 0.5698078 \\[6pt]
&= 0.9490532.
\end{aligned} \end{equation}$$
This is a large p-value and so we would not reject the null hypothesis in this case. Hence, there is no significant evidence to falsify the null hypothesis that $\sigma^2 = 9$.
A common alternative to the above test is to use a simpler calculation that takes the smallest of the two tail areas from above and below the observed test statistic, and then doubles this value. This is often used as a "quick-and-nasty" approximation to the above method, since it is relatively simple and does not require the calculation of a density cut-off value. (This is the method you are applying in your question.) Under this method, the p-value is approximated as:
$$\hat{p}(\chi_\text{obs}^2) = 2 \cdot \min \Big( \mathbb{P}(\chi^2 \leqslant \chi_\text{obs}^2 | H_0), \mathbb{P}(\chi^2 \geqslant \chi_\text{obs}^2 | H_0) \Big).$$
In your particular case this is an odd calculation, since the observed test statistic is above the mode of the null density, but its left-tail is the smaller probability. Your calculation of this approximate p-value is correct, but it is notable that it is not a very good approximation in this case (and in general, this is not a very good way to get p-values for a two-sided test with an asymmetric null distribution).
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