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Prove the hypothesis that the average content of containers of a particular lubricant is $10$ liters, if the contents of a random sample of $10$ containers are:

\begin{array}{|c|c|c|c|c|} \hline 10,2 & 9,7 & 10,1 & 10,3 & 10,1 \\\hline 9,8 & 9,9 & 10,4 & 10,3 & 9,8 \\\hline \end{array}

Use a significance level of $0.01$ and assume that content distribution is normal.

I started with the following:

In this case, use $t=\dfrac{\overline{x}-\mu_0}{s/\sqrt{n}}$, because I do not know anything about $\sigma$.

  • $H_0: \mu= 10$.
  • $H_1: \mu \neq 10$.
  • $\alpha = 0,01$.
  • How can I find the critical region?
  • How I can decide if the hypothesis is correct?

Thank you very much.

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1 Answer 1

up vote 1 down vote accepted

Your test statistic is $\dfrac{\bar x - 10}{s/\sqrt{10}}$. The "10" in the numerator comes from the null hypothesis; the "10" in the denominator is the sample size.

You need the number $c$ such that $\Pr(-c<T<c)=0.99$, when $T$ has a Student's t-distribution with $9$ degrees of freedom. In this case the number of degrees of freedom is one less than the sample size because if you subtract $\bar x$ from each observed number, you get a set of numbers that must add up to $0$; hence if you know $9$ of them, you know all of them. You can get that number $c$ from a table or from software. You reject $H_0$ if $T>c$ or $T<-c$, or equivalently, if $|T|>c$. That's the critical region.

The null hypothesis is presumed correct unless it is rejected, but you can never prove by such means that it is exactly correct.

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Thank you very much. –  Hiperion Nov 24 '12 at 17:29

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