02 40 205 8 08 3836 x z x so our conclusion is

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* .02 * 4.0 2.05 .8 100 4.0 2.05 .08 3.836 X z X = − = = × =
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So our conclusion is . . . Suppose our sample yielded a sample mean of 3.95 hours. Since 3.95 > 3.836, we would accept the null hypothesis.
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3.70 3.85 4.00 4.15 4.30 0 1 2 3 4 5 Normal Density Distribution of the Sample Mean when mu = 4 This is the picture. Accept Reject .02 3.836 3.95
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Other ways to do the test Once again, there are two other ways to perform the test. Compare observed z-values to critical z-value(s). Compare a p-value to α . Let’s repeat the test, using the z-value technique.
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