Answer:
The standard deviation of tire life does exceed 3000 km
Step-by-step explanation:
This is a Chi-Square Hypothesis Test for the Standard Deviation, here we have
![\bf H_0: \sigma = 3000](https://tex.z-dn.net/?f=%20%5Cbf%20H_0%3A%20%5Csigma%20%3D%203000)
![\bf H_a: \sigma > 3000](https://tex.z-dn.net/?f=%20%5Cbf%20H_a%3A%20%5Csigma%20%3E%203000)
so, this is an upper one-tailed test.
The <em>test statistic</em> is
![\bf T=\frac{(n-1)s^2}{\sigma^2}](https://tex.z-dn.net/?f=%5Cbf%20T%3D%5Cfrac%7B%28n-1%29s%5E2%7D%7B%5Csigma%5E2%7D)
where
<em>n = 10 is the sample size</em>
<em>s = sample standard deviation</em>
= population standard deviation
we would reject the null hypothesis if
![\bf T>\chi^2_{(\alpha,n-1)}](https://tex.z-dn.net/?f=%5Cbf%20T%3E%5Cchi%5E2_%7B%28%5Calpha%2Cn-1%29%7D)
where
![\bf \chi^2_{(\alpha,n-1)}](https://tex.z-dn.net/?f=%5Cbf%20%5Cchi%5E2_%7B%28%5Calpha%2Cn-1%29%7D)
is the critical value corresponding to the level of significance
with n-1 degrees of freedom.
we can use either a table or a spreadsheet to compute this value.
In Excel use
CHISQ.INV(0.05,9)
In OpenOffice Calc use
CHISQINV(0.05;9)
and we get this value equals 3.3251
Working out our T statistic
![\bf T=\frac{(n-1)s^2}{\sigma^2}=\frac{9*(2990)^2}{(3000)^2}=8.9401](https://tex.z-dn.net/?f=%5Cbf%20T%3D%5Cfrac%7B%28n-1%29s%5E2%7D%7B%5Csigma%5E2%7D%3D%5Cfrac%7B9%2A%282990%29%5E2%7D%7B%283000%29%5E2%7D%3D8.9401)
Since T > 3.3251 we reject the null and conclude that the standard deviation of tire life exceeds 3000 km.