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.
3.703.854.004.154.30012345Normal DensityDistribution of the Sample Meanwhen mu = 4This is the picture.AcceptReject.023.8363.95
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Other ways to do the testOnce 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.