Comment continued. I assume you are using normal data. It is not possible to work such a problem without some assumption about the population distribution. At an elementary level, the background should have been covered to allow answering for normal data. For example, the assumption of normal data leads directly to the chi-squared distribution of $Q$ below.
The significance level is the probability of rejection $P(S^2 \ge 13)$ under the assumption that $\sigma = 3.$ I guess your assignment is to get an answer using the relationship in my Comment above: With $Q = \frac{(n-1)S^2}{\sigma^2},$ you have
$Q \sim \mathsf{Chisq}(\nu),$ where $\nu = n-1 = 29-1 = 28$
degrees of freedom.
For your data with $n = 29, S^2 = 16$ and your null hypothesis $\sigma = 3,$ you have
$$Q = \frac{(n-1)S^2}{\sigma^2} = \frac{28(16)}{9} = 49.78.$$
Now, the critical value for a test is given as $S^2 > 13.$
But the printed chi-squared table is given in terms of $Q,$ which is the chi-squared random variable. So $S^2 > 13$ is equivalent
to
$$Q_c = \frac{28S^2}{9} = \frac{28(13)}{9} = 40.44.$$
So you reject $H_0,$ according to the stated critical region, either because $S^2 = 16 > 13$ or because $Q = 49.78 > 40.44.$
If you look at the row for DF = 28 in your printed chi-squared
table, you will find that 49.78 lies between the entries 48.27 and 50.99, in columns headed .01 and .005., indicating that the values cut about 1% and 0.5% from the upper tail of the relevant chi-squared distribution. So I get that the p-value lies between
1% and 0.5%, not between 5% and 0.25% as in the answer provided. The exact
p-value is 0.007, which lies between 0.01 and 0.005.
qchisq(c(.9, .95, .975, .99, .995), 28)
[1] 37.91592 41.33714 44.46079 48.27824 50.99338
1 - pchisq(49.78, 28)
[1] 0.006842861
The critical values implied by $S^2 \ge 13$ or $Q > 40.44$ are
not for a test at the 5% significance level.
The level based on those critical values is about 6%.
1 - pchisq(40.44, 28)
[1] 0.06038755
A test at
the 5% level would use $Q \ge 41.34$ or $S^2 = 9(41.34)/28 \approx 13.29.$
For verification, a simple simulation in R gets approximately these same results (6%, 13.3, 0.007), directly from the simulated distribution of $S^2,$ when $n = 29$ and $\sigma = 3.$
set.seed(1214); m = 10^5; n = 29; sg = 3; mu = 8
v = replicate(m, var(rnorm(n, mu, sg)))
mean(v > 13)
[1] 0.06023
quantile(v, .95)
95%
13.26989
mean(v > 16)
[1] 0.00683
The figure below shows a histogram of the simulated values of $Q$ with the
density function of $\mathsf{Chisq}(28).$ The heavy vertical black line is at the
observed value of $Q$ corresponding to $S^2 = 16.$ Thin vertical brown and cyan
lines cut probabilities 5% and 1%, respectively, from the upper tail of the distribution.
So I don't get the alleged answer. I don't know whether we are dealing with a misinterpretation of the problem, with typgraphical/computational errors, or with an incorrect answer. Please check everything.
Notes: (1) In R statistical software the function pchisq denotes the CDF of a chi-squared distribution and the function qchisq indicates its inverse CDF or 'quantile' function. (2) In the simulation v is a vector of 100,000 simulated values of $S^2.$
