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Unformatted text preview: ase 1 σ known RightTailed Test Example Suppose now that you also believe that µ > 1.5. So you want to test the claim µ ≤ 1.5. Let’s do
that. First, let’s properly state the hypotheses:
H0 :µ ≤ 1.5
H1 :µ > 1.5
This is a righttailed test. Under the assumption that null hypothesis is true, we need to evaluate
the chances of x > 2.5. If small, the null hypothesis is not likely to be true.
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Next step: relevant test statistic is z = x −µ0
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√
σ/ n = 2.5−1.5
0.3 = 3.33. Next: Critical value for α = 0.01 is z0.01 = 2.33.
Next: Use the decision rule: z > 2.33 so we reject H0 .
OR: pvalue for z = 3.33 is smaller than 0.001, so we reject the null hypothesis for reasonable α.
Exercise: Calculate the smallest x value that leads to rejection of the null.
¯ Utku Suleymanoglu (UMich) Hypothesis Testing 17 / 39 What determines test results? Determinants of The Test Results Before we go on and discuss twotailed testing and other cases, let’s talk about a few things:
Diﬀerence between hypothesized parameter and calculated statistic: x − µ0 . Generally
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speaking, if the claim is too far oﬀ from the hypothesized value, test statistic would be larger in
absolute terms. The eﬀect of this on test depend on the sign of the test statistic and the tail of
the test.
Here is an example: Remember we have x = 2.5 and σ = 1.5 and n = 25. Suppose now that you
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have to test H0 : µ ≥ 1 with the alternative hypothesis H1 : µ < 1.
We can calculate the test statistic: z = 2.5−1
0.3 = 5. That is a pretty big z . But this is a lefttailed test. The pvalue is calculated as the left tail probability. P (Z ≤ 5). It is
almost 1, it is bigger than any α imaginable. So you fail to reject the null.
Rejection region is (∞, −1.645) (if α = 0.1). Critical value is negative for lefttailed tests, so a
positive test statistic cannot be in the rejection region.
Even if the test results seems obvious, we still test it properly. And this is an example why a
highe...
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 Spring '08
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