Hello wta2k!
A one-tail test at the 0.05 level of significance for this exercise
means the following:
If the true mean of the distribution is $2.50, what is the probability
of observing a sample mean of $2.10 or less? If this probability is
sufficiently low (in this case, lower than 0.05), then you would
reject that the true mean is $2.50. That is the spirit of a one-tail
test of hypothesis. It's crucial that this is clear. The null
hypothesis is that the population mean is $2.50. The alternative
hypothesis is that it is lower than that.
So, to determine wether this experience suggests a decline in
spending, you first have to find the probability of observing the
sample mean of $2.10 or less. In order to do this, you use the fact
that if you take the sample mean from a normal distribution, then the
sample mean itself is normally distributed. In particular, the sample
mean has a normal distribution with mean equal to the population mean,
and standard deviation (SD) equal the population SD divided by the
square root of the number of observations.
In this exercise, you would have to determine Prob(Xbar < 2.10) where
Xbar is the sample mean, which you know has a normal distribution,
with mean 2.5 (under the null hypothesis) and standard deviation
0.9/sqrt(18), where sqrt(18) means the square root of 18.
In order to compute the probability Prob(Xbar < 2.10), tables in
textbooks are generally used. You can also find it, for example, at
http://www.math2.org/math/stat/distributions/z-dist.htm
As you will notice, these tables state Prob(X < a) for any number a,
but with X being a *standard* normal distribution (that is, with
mean=0 and SD=1). So, you need to "convert" Xbar into a standard
normally distributed variable. To do this, you substract the mean and
divide by the SD, in the following way:
Prob(Xbar < 2.1)
=Prob(Xbar - 2.5 < 2.1 - 2.5)
=Prob( (Xbar-2.5)/s < (2.1-2.5)/s )
where "s" is the SD, that is, s=0.9/sqrt(18). This amounts to
=Prob( (Xbar-2.5)/s < -1.88 )
Also, now (Xbar-2.5)/s is standard normally distributed, so you can
now use the table. You will find that Prob( (Xbar-2.5)/s < -1.88 ) =
0.03005. I got this number by checking the link I provided above. Look
at row -1.8, then go to column 0.08 (columns are the second decimals).
Voila! The probability that you observe a sample mean of 2.1 if the
true mean is 2.5, i.e. 0.03005, is the p-value. How do you interpret
this number? It means that there is only a 3% probability that you
observe a $2.10 sample mean if the population (true) mean is $2.50.
Therefore, given that you have actually observed the $2.10, it's
"unlikely" that the population mean is $2.50. Thus, you *reject* the
hypothesis that the population mean is 2.5. The publicity campaign
really did have an effect.
What does the 0.05 level of significance have to do with this? Well,
setting the level of significance at 0.05 means that if you had
obtained a p-value of 0.05 or higher (a probability of 5% or higher
that sample mean 2.1 is observed with true mean equal to 2.5) then you
would have NOT rejected the null hypothesis. The level of significance
sets what is "unlikely" and what isn't.
I hope the explanation was clear enough. If you have any further
questions please post a clarification request prior to rating the
answer. Otherwise, I welcome your rating and your final comments and I
look forward to working with you again in the near future. |
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