Hello!
Here are your answers.
Question 1
The procedure to solve questions regarding hypothesis testing on the
population mean is the following. First of all, you need to identify
whether you're using a one- or a two-tailed test. This depends on the
form of the alternative hypothesis: for example, if it's u<90, or
u>90, then it's a one-tailed test. If it's u [not =] 90, then it's a
two-tailed test. Clearly, this first question asks about a one-tailed
test.
Given the null and alternative hypothesis, and the significance level
(in your case, 0.05), we need to find the rejection region. This will
be an interval such that if the observed sample mean lies within it,
then it will be "safe" (a 95% confidence level in this case) to
conclude that the null hypothesis is false and reject in favor of the
alternative hypothesis.
In this case, the null is u=90 the alternative hypothesis is u>90.
Therefore, we would reject the null hypothesis if the actual sample
mean were "much greater" than 90, and assume that the null hypothesis
is true. Specifically, the rejection region, will "start" in a number
(let's call it 'a') such that if the actual mean were really 90, then
the probability of observing a sample mean greater than 'a' would be
0.05.
So let's write the math for this. Let's call Xbar to the observed
sample mean. We want find 'a' such that:
Prob (Xbar > a) = 0.05
We don't know the distribution of Xbar in order to solve this.
However, we can make use of the Central Limit Theorem. This theorem
states that, for a large sample size (usually more than 30
observations),
(Xbar - m)/ (s/sqrt(N))
follows approximately a standard normal distribution; where m is the
population mean, s is the population standard deviation and sqrt(N) is
the square root of the sample size.
In your case, we know that s = 22.5494... and sqrt(N) = 10; so
s/sqrt(N) = 2.25494...
Now, using this theorem, let's "convert" the Xbar in our equation into
a standard normal distribution. Recall we're using that the population
mean is actually equal to 90, so m=90:
Prob (Xbar > a) = 0.05
Prob (Xbar - 90 > a - 90) = 0.05
Prob ( (Xbar - 90)/2.25494 > (a - 90)/2.25494 ) = 0.05
Given the Central Limit Theorem, we now know that the left-hand side
of the inequality follows a standard normal distribution. Let's rename
it as Z:
Prob ( Z > (a - 90)/2.25494 ) = 0.05
We can now solve this equation. Looking up in a standard normal
distirbution table (which can be found in most Statistics books), we
get that:
Prob(Z > 1.64) = 0.05
[Please do let me know through a Clarification Request if you don't
know how to use these tables]
This 1.64 is the critical z-score. If the test statistic (which we'll
see how to compute in a few moments) is greater than this z-score,
then we'll reject the null hypothesis. If it's smaller, then we won't
reject the null hypothesis. This critical z-score can also be used in
the following way. We found that:
Prob ( Z > (a - 90)/2.25494 ) = 0.05
Prob(Z > 1.64) = 0.05
Therefore,
(a - 90)/2.25494 = 1.64
a = 1.64*2.25494 + 90
a = 93.69...
This implies that if the observed sample mean is greater than 93.69,
then we'll reject the null hypothesis. In this case, we're told that
that the sample mean is 94.4. We thus REJECT THE NULL HYPOTHESIS.
We can also use test statistic in order to arrive to the same
conclusion. It's calculated as:
(Observed Sample Mean - Mean from Null hypothesis)/ (s/sqrt(N))
= (94.4 - 90) / 2.25494
= 1.9512
As you can see the test statistic is greater than the critical
z-score, therefore, we reject the null hypothesis.
Question 2
The logic in this test is very similar to the previous one, with the
difference that we now need to use a two-tailed test. As such, the
rejection region will now be "spli in two": we'll reject the null
hypothesis if the observed sample mean is "too large" but also if it's
"too small".
We now want would like two values 'a' and 'b' such that:
Prob(Xbar > a) = 0.025
Prob(Xbar < b) = 0.025
The 0.025 comes from the fact, in two-tailed tests, we need to divide
alpha (0.05) by 2, because there are two "separate" rejection regions.
By doing this, the sum of the probabilities in the two rejection
regions will be 0.05, which is the desired significance level.
We now proceed exactly as we did in the previous question. We'll start using
Prob(Xbar > a) = 0.025
We "convert" it to standard normal just as we did before:
Prob( (Xbar - 90)/2.25494 > (a - 90)/2.25494 ) = 0.025
Prob( Z > (a - 90)/2.25494 ) = 0.025
Looking up in a table, we get that:
Prob( Z > 1.96 ) = 0.025
Therefore, the POSITIVE CRITICAL Z-SCORE IS 1.96. Alternatively, we
can say that the "upper" rejection region are the values for the
sample mean greater than 90 + 1.96*2.25494 = 94.42. Since the sample
mean is smaller than this level (94.4) we still can't reject the null
hypothesis - we'll arrive to the same conclusion later with the test
statistic.
Let's now calculate the other rejection region. We had:
Prob(Xbar < b) = 0.025
Doing the same calculations, we get:
Prob( Z < (b - 90)/2.25494 ) = 0.025
We don't need to use a table for this. Since the standard normal
distribution is symmetric around zero, it has the following property:
Prob ( Z > x) = Prob( Z < -x)
Therefore, since the probability is 0.025 in both cases, then we
conclude that the critical z-score from this equation is the opposite
of the one we found earlier. Specifically, the NEGATIVE CRITICAL
Z-SCORE IS -1.96. We can also compute the lower rejection region in
terms of the sample mean just as we did before:
(b - 90)/2.25494 = -1.96
b = 90 - 1.96*2.25494 = 85.58
This means that if the sample mean were lower than 85.58, we would
reject the null hypothesis. Since it's neither lower than 85.58 nor
greater than 94.42 (it's 94.4), then we conclude that we CANNOT REJECT
THE NULL HYPOTHESIS.
Let's find the same solution with the test statistic. This statistic
is calculated in excatly the same way as before:
(Observed Sample Mean - Mean from Null hypothesis)/ (s/sqrt(N))
= 1.958
Since the test statistic is neither greater than the positive critical
z-score, nor lower than the negative critical z-score, then we
conclude that we CANNOT REJECT THE NULL HYPOTHESIS.
I hope this helps! If you have any doubt regarding my answer, please
don't hesitate to request clarification before rating it. Otherwise, I
await your rating and final comments.
Best wishes!
elmarto |
