A engineer conjectures that the population mean of a certain component parts is 5.0 millimeters. An experiment is conducted in which 100 parts produced by the process are selected randomly and the diameter measured on each.

It is known that the population standard deviation σ = 0.1. The experiment indicates a sample average diameter X = 5.027 millimeters.

Does this sample information appear to support or refute the engineer’s conjecture?

The solution goes by this

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here is what bothers me, why is that, he considered the area P(z>2.7). From my point of view, it should be set up like this P(z<2.7) (since we are finding the probability that the length should be between 0 and 0.027 mm. Then it yields 0.99 which means, 99% of the time, it falls with -0.027 - 0 - 0.027 and should support the engineers conjecture. Thanks

Please illuminate me if Im wrong

  • 3
    $\begingroup$ With this reasoning, if the sample average diameter is X=42 kilometers, this supports even more strongly the hypothesis that the actual mean is 5.0 millimeters... Fishy, don't you think? $\endgroup$ – Did Mar 2 '12 at 7:10

There are situations where there can be disagreement about the proper Null Hypothesis and Alternate Hypothesis. However, in this case the engineer's Null Hypothesis is quite clear from the wording of the problem: the mean is $5.0$.

The default Alternate Hypothesis is that the mean is not equal to $5.0$. If another Alternate Hypothesis was to be used, there should be some wording to suggest it, maybe that the engineer fears the mean may be greater. In any case, the Alternate Hypothesis needs to be formulated before measurements are taken.

Your calculations suggest that it is reasonable to reject the Null Hypothesis. The number $0.007$ is, however, not overwhelmingly small. It might be reasonable to test a larger sample.

I do not know precisely what you are arguing. On the Null Hypothesis, the probability that the sample mean is $<5.027$ is about $0.9965$, so the probability of getting a result as high as $5.027$ or higher is only $0.0035$, half of your $0.007$. I do not see where the $0.99$ comes from.

  • $\begingroup$ im arguing why is tht he used greater than symbol. it should be less thn $\endgroup$ – WantIt Mar 2 '12 at 7:48
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    $\begingroup$ @blackandyellow: He was trying to find out how likely it is that the sample mean would be $0.027$ away from $5.0$, assuming, as is always done in hypothesis testing, that the null hypothesis is true. He got the result $0.007$, which was interpreted as very unlikely. So, assuming the null hypothesis, something quite odd happened, which is why he rejected the null hypothesis. $\endgroup$ – André Nicolas Mar 2 '12 at 8:03

I can only write a solution, not a comment, I guess since I don't have enough points yet.

Basically, the issue is that your (representative ) sample is 2.7 deviations from the alleged true value of 5. Under the assumed distribution (normal,$\mu=5$, $\sigma=0.1$, this happens only 0.007% of the time.

Basically, there is such a thing as natural variation: even if the true mean is 5, there is enough "noise" that would push your measurements of this exact value. But, if the variation is due just to noise, it is not likely to be $2.7 \sigma's$ from the mean; actually, your calculations show that a deviation of this sort is very rare, so we conclude that it is not likely due to chance alone, and it is more likely that the true populat


With this kind of hypothesis testing, the convention is to see whether the probability that the observation or something more extreme occurs (assuming the null hypothesis) is low, and if so reject the null hypothesis.

This means that if the observation is more extreme, the probability of it or something even more extreme occurring by chance is also smaller, making you more confident in your rejection of the null hypothesis; as Didier Piau says in his comment, your suggested approach would instead lead to extreme observations making you more confident that the null hypothesis was true.

Part of the confusion may come from so-called "99% confidence intervals" being wider than "95% confidence intervals". In fact, what you are 95% or 99% confident of is that this method should mean you will not reject the null hypothesis by chance when it happens to be true with that probability. This tells you little explicit about the probability of the null hypothesis actually being false and is such a tangle of negatives and hypotheticals that some people reject this kind of description altogether and move to Bayesian methods.


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