## Related questions with answers

Find the test statistic and critical value(s). Also, use the mentioned table to find limits containing the P-value, then determine whether there is sufficient evidence to support the given alternative hypothesis.

Precipitation Amounts $H_1: \sigma \neq 0.25, \alpha=0.01, n=26, s=0.18$.

Solution

Verified$H_0:\sigma =0.25$

$H_1:\sigma \neq 0.25$

Compute the value of the test statistic:

$\chi^2=\dfrac{n-1}{\sigma^2}s^2=\dfrac{26-1}{0.25^2}\cdot 0.18^2=12.96$

Determine the critical value(s) using table A-4 with $df=n-1=26-1=25$:

$\chi^2_{1-0.005}=\chi^2_{0.995}=10.520$

$\chi^2_{0.005}=46.928$

The P-value is the interval that contains the critical value in table A-4:

$0.02<P<0.05$

If the test statistic is in the rejection region, then reject the null hypothesis:

$10.520<12.96<46.928\Rightarrow \text{ Fail to reject } H_0$

There is no significant evidence to reject the claim.

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