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There was too much snow on the highways, so the mayor of the town sent snowplows to spread some chemicals on them. There is a standard of how much of one specific substance should be present in the compound that is used for spreading... We measured how much of the substance was present in the compound in 30 different places of the town. These are the results:

0.91 1.08 0.72 1.07 1.14 0.62 1.06 1.20 0.76 1.19   
0.96 0.73 0.83 0.55 0.79 1.34 0.60 1.19 1.35 1.13   
0.67 0.77 0.48 0.83 1.78 2.25 1.21 0.89 0.83 1.07

We expect that the values have normal distribution. Verify with a reliability of 99% that the standard deviation is less than 0.4. [Result: r = 24.546. Hypothesis H0 is not denied.]

I calculated

a) $\mu$ = 1.00.....and.....b) $\sigma$ = 0.367

Now I set ...H0: $\sigma^{2} = \sigma^{2}_{o}$... versus...H1: $\sigma^{2} < \sigma^{2}_{o}$

I used this test: $ \frac {(n-1) s_{n}^{2}} { \sigma_{0}^{2} } \leq \chi^{2} _{ \alpha } (n-1) $

Then, I calculated $ \frac {(n-1) s_{n}^{2}} { \sigma_{0}^{2} } $ = 24.54 and $\chi^{2} _{ \alpha } (n-1) $ = 49.58

Now, we see that the inequality holds good, so H0 should be denied! However the result in the book says the opposite...

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  • $\begingroup$ The associated p-value is greater than 1%, therefore there is not enough evidence to reject the null hypothesis. Note that the statistic is smaller than the critical value. I have noticed that you got very good answers in the past and you seem to be satisfied, consider accepting some of them to motivate people to continue to help you. It is just one click ;) $\endgroup$
    – user10525
    Jun 10, 2012 at 12:03
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    $\begingroup$ I didn't know there was something like Accepting Answers. From now on, I will always accept the best answer to my question which has at least 1 answer. Thank you Procrastinator for letting me know! $\endgroup$ Jun 10, 2012 at 12:59

1 Answer 1

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As Procrastinator pointed out the test statistic is not significantly large. Don't just look at the number and assume that it is large enough to reject! The chi square statistic has 29 degrees of freedom. It has a mean of 29 and a variance of 58. So the value of the test statistic being 24.54 is not large at all and with the estimate so close to 0.4, this is what we would expect.

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  • $\begingroup$ So when do I use p-value as a test and when Normal Distribution Parameters Testing? $\endgroup$ Jun 10, 2012 at 12:13
  • $\begingroup$ When doing hypothesis testing you compute the test statistic and compare it to the critical value to determine whether or not to reject the null hypothesis. The p-value add information by telling how extreme your result would be if the null hypothesis were true. A p-value greater than 0.05 is customarily taken as not significant. In this case Procrastinator checked to find the p-value to be greater than 0.2 $\endgroup$ Jun 10, 2012 at 12:23
  • $\begingroup$ So, is it like this? -> .... When I get the result from the test that I should reject H0 and H1 is true, I am obliged to calculate p-value and then according to the p-value decide if the result is significant or if it's not. On the contrary, when the result of the test says that I cannot reject H0 because the inequality doesn't hold good, I don't have to calculate p-value. Or do I have to calculate p-value all the time? $\endgroup$ Jun 10, 2012 at 12:53
  • $\begingroup$ To test a null hypothesis you only need to compare the test statistic to the critical value. You never have to compute the p-value. The p-value just provides more information than just a statement of reject/ don't reject. You can always compute a p-value. There is no inconsistency. In your case the p-value was much higher than 0.05 and your test statistic was well below the critical value. The p-value is more interesting when you reject because a p-value of 0.001 provides a clearer sign that you should reject than say a p-value of 0.03. $\endgroup$ Jun 10, 2012 at 12:53
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    $\begingroup$ My last question: How do I calculate p-value in this case? $\endgroup$ Jun 10, 2012 at 13:30

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