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In this problem, the calories per gram are normally distributed with mu equals 100 and sigma equals 2 at the mean and standard deviation respectively.
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We want to test the null hypothesis h0 at mu equals 100 against the alternative of mu does not equal 100 with a sample of size n equals 5.
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Questions a through c below relate to the nature of hypothesis testing and how to evaluate the general framework of an implement hypothesis test.
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Question a says that the exception region is x between 98 .5 and 101 .1 .5.
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What is alpha the probability of a type 1 error? to reduce alpha, we utilize the bounds of our critical region relative to our mean.
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So, for a normal variable, we can deduce the z -not as follows.
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Z -0 is x -1 -new over sigma root n.
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If we choose the lower bound, this is 98 .5 minus 100 over 2 by 98 .5.
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D -9 .67, and the probability z less than z -not is 0 .075 from a normal distribution.
00:57
This is alpha over 2 because our test is a two -tailed test, therefore alpha is split between the two tails...