Hi again!
Finally, here are the answers to your questions.
- What would be the probability that X-bar will be within 2 units if
the
random sample n = 10 and population mean = 67 and SD = 3.24?
Following your clarification, we assume the population is normally
distributed. Furthermore, we know exactly the sampling distribution of
the sample mean (Xbar): Xbar is normally distributed, with mean 67 and
SD=3.24/sqrt(10)=1.02 (sqrt means "square root of"). You can check the
following link to see why this is so.
Sample Means
http://www.stat.yale.edu/Courses/1997-98/101/sampmn.htm
Now, we want to calculate
Prob( |Xbar-67| < 2 )
This is the probability that Xbar is within two units of the
population mean, which is 67. Using the symmetry property of the
normal distribution, we can see that calculating this probability is
exactly the same as calculating
1 - 2*Prob( Xbar-67 > 2)
So let's focus on the term Prob( Xbar-67 > 2)
We can rewrite this as
Prob( (Xbar-67)/1.02 > 2/1.02 )
=Prob( Z > 1.96)
I rewrote the lefthand side of the inequality as Z because that term
follows a standard normal distribution. This is so because Xbar
follows a normal distribution with mean 67 and SD 1.02. If you take a
normally distributed random variable, substract its mean and divide it
by its SD, you get a standard normally distributed random variable. So
now we're ready to use a standard normal distribution table, such as
the one in the following link.
Normal Distribution Table
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/normaltable.html
Looking up 1.96 in the table, we get that
Prob(Z>1.96) = 0.025
Therefore,
Prob( |Xbar-67| < 2 )
=1 - 2*Prob( Xbar-67 > 2)
=1 - 2*0.025
=0.95
So 0.95 is the probability that the sample mean (Xbar) falls within
two units of the population mean.
- From 15 observations of normal population randomly sampled, Xbar =
150 and S=10. If the level of significance is set at 0.01, how can
hypothesis
Ho: m = 160 tested against H1:m <160?
Using the same reasoning as before, we know that Xbar is normally
distributed, with mean equal to the population mean (unknown, let's
call it 'm') and SD=10/sqrt(15)=2.58. I have assumed here that the
S=10 you provide in the question is the *population* SD and not the
sample SD. If this is not correct, please let me know through a
clarification request.
The idea of the hypothesis test is the following. If Xbar's true mean
were 160 (the null hypothesis) we would expect to observe a value of
Xbar not very different from 160. The observed value is 150. Is this
difference "large enough" to reject the hypothesis that the true mean
is 160? This what we test with hypothesis testing. We have to find the
probability of observing a value for Xbar of 150 or less given that
Xbar's mean is 160. If this probability is too small, we conclude that
it's "unlikely" that Xbar's true mean is 160; and thus we reject the
null hypothesis. Otherwise, we can't reject it. How "small" this
probability should be in order to reject the null hypothesis is a
threshold that must be given in the question. In this case, you
specified 0.01 (the significance level). Therefore, if the probability
of observing 150 or less is less than 0.01, we'll reject the null
hypothesis.
So we must first compute the probability of observing 150 or less
given that Xbar follows a normal distribution with mean 160 and SD
2.58. We proceed just as before:
Prob(Xbar =< 150)
=Prob(Xbar-160 <= 150-160)
=Prob((Xbar-160)/2.58 <= -10/2.58)
=Prob( Z <= -3.87)
Using the symmetry property, this is the same as calculating
Prob(Z >= 3.87)
Again, looking up the table we find that
Prob(Z >= 3.87) = 0.001
which is smaller than 0.01. Therefore we reject the null hypothesis
that m=160. We have found that there's only a 0.001 probability that
Xbar is 150 or less given that the population mean is 160; so this
premise (m=160) appears to be wrong.
Google search strategy
hypothesis testing
://www.google.com.ar/search?q=hypothesis+testing&ie=UTF-8&oe=UTF-8&hl=es&meta=
normal distribution table
://www.google.com.ar/search?hl=es&ie=UTF-8&oe=UTF-8&q=normal+distribution+table&meta=
I hope this helps! If you have any doubt regarding my answer, please
request a clarification before rating it. Otherwise I await your
rating and final comments.
Best wishes!
elmarto |
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