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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?

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  • $\begingroup$ Actually, I got $\chi^2_{0.99}(40)=22.1643$. $\endgroup$ Commented Oct 3, 2018 at 23:34
  • $\begingroup$ @WinterSoldier : You're putting probability $0.99$ in the upper tail and $0.01$ in the lower tail. The poster is doing it the other way around. $\endgroup$ Commented Oct 3, 2018 at 23:35
  • $\begingroup$ @MichaelHardy, yes you are correct. $\endgroup$ Commented Oct 3, 2018 at 23:38
  • $\begingroup$ @WinterSoldier I got the same number initially. And I did it the second time using an online calculator to derive the answer. I got 29.051 the second time. Am I right? $\endgroup$ Commented Oct 3, 2018 at 23:43
  • $\begingroup$ @Lady : See the comment under my answer. $\endgroup$ Commented Oct 4, 2018 at 3:50

2 Answers 2

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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.

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  • $\begingroup$ when I used this link (stattrek.com/online-calculator/chi-square.aspx) the critical value is 29.1? Does it mean I am wrong? $\endgroup$ Commented Oct 3, 2018 at 23:54
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    $\begingroup$ @Lady: $$\begin{align} & \Pr(\chi^2_{40} < 22.16426) = 0.01 \\ \\ & \Pr(\chi^2_{40} > 63.69074) = 0.01 \\ \\ & \Pr(\chi^2_{40} < 29.05052) = 0.1 \end{align}$$ It seems you're looking at the probability in the lower tail rather than in the upper tail, and you used $0.1$ instead of $0.01. \qquad$ $\endgroup$ Commented Oct 4, 2018 at 3:49
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$63.691$ 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.

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  • $\begingroup$ when I used this link (stattrek.com/online-calculator/chi-square.aspx) the critical value is 29.1? Does it mean I am wrong? $\endgroup$ Commented Oct 3, 2018 at 23:53
  • $\begingroup$ I think you must have input 0.1 instead of 0.01. Use a table instead. $\endgroup$ Commented Oct 4, 2018 at 0:01
  • $\begingroup$ When using a table, one can see the trend in dfs and $X^2$ values versus probabilities. $\endgroup$ Commented Oct 4, 2018 at 0:09

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