Hello again wta2k!
I've already answered your question #185198. You may want to read that
one first, since I use some concepts that are explained in that
question.
This one is similar in some ways, but with two important differences:
a two-tail test should be used in this case, and we don't know the
distribution of the length of the nails. The second problem is
addressed easily: there is a theorem, called the "Central Limit
Theorem" that states that for large samples, the sample mean has
approximately a normal distribution, even if the population where the
sample mean came from is not normal. In most cases, 30 observations do
for a sample to be "large". So, with 35 observations we can be pretty
sure that we can treat the sample mean as being normally distributed
even if it is not, and we'll still get quite accurate results.
In a two-tail test, we don't want the sample mean to go away from the
population mean in *both* directions: we don't want it to be either to
high or too low. Thus the 0.01 of the level of significance is divided
equally between both tails, thus we will reject the null hypothesis in
2 cases:
1) if the probability of observing a sample mean equal or less than
the one we observed is smaller than 0.005
2) if the probability of observing a sample mean equal or greater than
the one we observed is smaller than 0.005
The 0.01 is level of significance is "divided" in this way in
0.005+0.005. Note how this is different from the other question: in
the other one, we only cared for observing a very small sample mean;
we didn't mind if the sample mean was way higher than the population
mean. That's the basic difference between a one-tail and a two-tail
test.
Now, using the same procedure as in question #185198, we find that
Prob( (Xbar-2)/s > (2.025-2)/s
=Prob( (Xbar-2)/s > 2.11 )
=0.01743
(recall that s is the standard deviation divided by the square root of
the number of observations, in this case s=0.011)
So, finding that Prob( (Xbar-2)/s > (2.025-2)/s ) equals 0.01743
(which is not smaller than 0.005) means that for now we can't reject
the null hypothesis that the population mean is 2.000 (i.e., that the
machine is adjusted properly). What about checking the probability of
observing an Xbar equal or *lower* than the one observed? Given that
the observed sample mean is higher than the assumed population mean,
it's clear that this probability will be high. You're looking at the
probability of observing a sample mean of less than 2.025, when the
mean is 2.000; it will not be small. Formally,
Prob( (Xbar-2)/s < 2.11 )
=1 - Prob( (Xbar-2)/s > 2.11 )
=1 - 0.01743
=0.9826
0.9826 is greater than 0.005. So, in conclusion, you can't reject the
null hypothesis that the machine is adjusted properly. What about the
p-value? In this case, since it's a two-tail test, the p-value is
2*0.01743 = 0.03486.
Again, since the p-value is greater than the level of significance of
the test (0.03486>0.01), we can't reject the null hypothesis. It maybe
noteworthy that with this p-value, if the investigator had chosen a
higher level of significance (such as 0.05), the null hypothesis would
have been rejected.
For additional information, you might want to check the following link
and hypothesis test
http://www.pinkmonkey.com/studyguides/subjects/stats/chap8/s0808g01.asp
I hope this helped. I had a good time working on these questions, I
really hope that the answers satisfy you. Again, if anything is
unclear to you, please request a clarification before rating this
answer.
Best regards,
elmarto |
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