chi-square distribution and p-value

Question

A study is conducted to test the hypothesis that people with glaucoma have higher variability in systolic blood pressure(SBP). The study includes 41 people with glaucoma whose mean SBP is 140 mmHg with a standard deviation of 25 mmHg. If the population standard deviation is 20 mmHg, verify the claim at 1% significance level. Also provide the p-value of the test statistic.

My attempt:

My null hypothesis is that $$ H_0: \sigma = \sigma_0 $$ My alternative hypothesis is that $$ H_a: \sigma > \sigma_0 $$

$$ C = \frac{(n-1)S^2}{\sigma^2} = \frac{(41-1)(25)^2}{20^2} = 62.5$$

I am bit confused on what my critical value should be. For a one-sided alternative, shouldn't the critical value be $\chi_{40,0.99}^{2} > 62.5$. My critical value is $\chi_{40,0.99}^{2} = 63.691$ which is greater than $62.5$. My p-value is $0.01295>\alpha=0.01$. How do I make my conclusion?

Answer 1

The only mistake I find in your posting is your assertion that $63.691<62.5.$

Since the value of the test statistic is less than the critical value, the null hypothesis is not rejected.

Likewise, since the p-value is more than $0.01$ the null hypothesis is not rejected.

Answer 2

$63.691$ is "greater" than $62.5$ not less which means your p value test statistic is $> 0.01$. Hence you fail to reject the null and conclude that there is no statistically significant difference in SBP variability and hence no evidence for the alternative hypothesis.