Question 1
In order to test this hypothesis, we need to do the following. First of all, we have to find
to find a confidence interval around the mean proposed by the null hypothesis. The
confidence interval is built with respect to the significance level, so let's start with the
case in which the required significance level is 5%. The confidence interval will then be a
set of values in which the sample mean could fall with 95% probability if it were true that
the population mean is 165.
After this we will be able to check if we can reject the null hypothesis. For example, let's
say that we find this confidence interval to be [162, 168]. This will mean that, given a
population mean of 165, random samples of 64 students will yield a sample mean that
will be in that interval with 95% probability. Therefore, if the actual mean we observe
(170 in this case) is outside that interval, we can reject the null hypothesis, because if the
population mean were truly 165, then the sample mean should have fallen inside that
interval almost surely (95% prob). Let's see how to find the confidence interval.
Let's call Xbar to the sample mean and s to the to find values 'a' and 'b' such that:
Prob ( Xbar > 165 + a ) = 0.025
Prob (Xbar < 165 - b) = 0.025
[that is, the probability that we're outside the interval [165-b, 165+a] is 0.05]
Now, given that the population is normally distributed, the distribution of the sample
mean (Xbar) will also be normal. In particular, the sample mean taken from a normal
population has mean equal to 'm' and standard deviation equal to s/sqrt(n) [sqrt() means
square root of], where m is the population mean, s is the population standard deviation
and n is the sample size. Although, we don't know the population mean (that's what we're
trying to test), we're going to assume it's equal to 165, the null hypothesis. Afterwards we
will be able to check if that assumption was a good one. Therefore, we know that Xbar
will have a normal distribution with mean 165 and standard deviation equal to 16/sqrt(64)
= 16/8 = 2.
Let's make a further manipulation in the equations we wrote above:
Prob ( Xbar - 165 > a ) = 0.025
Prob (Xbar - 165 < -b) = 0.025
Since the normal distribution is a symmetric around its mean (165), it's clear that 'a' and
'b' will be the same number. So we can just find one of them:
Prob ( Xbar - 165 > a ) = 0.025
Lastly, we'll use one more property of the normal distribution. Recall that when we take a
normally distributed variable, subtract its mean, and divide it by its standard deviation,
we get a standard normally distributed random variable; that is, a normal distribution with
mean 0 and std dev 1. So let's divide both sides of the inequality in the above equation by
the std dev of Xbar (which is 2):
Pob ( (Xbar - 165)/2 > a/2 ) = 0.025
Since we know that (Xbar - 165)/2 follows a standard normal distribution, let's replace
that term with Z (the symbol used to denote a standard normal distribution) to keep things
tidy:
Pob ( Z > a/2 ) = 0.025
Checking a standard normal distribution table (which can be found in most Statistics
books) we find that:
Prob (Z > 1.96) = 0.025
[obviously, this is the same value that appears in the table that you provided, since this is
a two-tailed test at the 0.05 significance level]
Therefore, we conclude that:
a/2 = 1.96
a = 1.96*2
a = 3.92
Therefore, the confidence interval is [165-3.92, 165+3.92] = [161.08, 168.92]. Since the
observed value of the sample mean (170) is outside this interval, we conclude that it's
significantly different from 165, so we reject the null hypothesis at the 0.05 significance
level, in favor of the alternative hypothesis (the population mean is different from 165).
As you've seen, the general formula for the confidence interval of a two-tailed test is:
Lower bound: (Mean from null hypothesis) - (Zvalue * Std Dev of Xbar)
Upper bound: (Mean from null hypothesis) + (Zvalue * Std Dev of Xbar)
For the case in which the significance level is 0.01, we can repeat the same procedure, or
directly use the formula I've just provided to get that:
Lower bound = 165 - (2.58 * 2) = 159.84
Upper bound = 165 + (2.58 * 2) = 170.16
[I've used the z-value provided in your table for two-tailed tests at the 0.01 significance
level]
The confidence interval is thus [159.84, 170.16]. Since the observed mean falls inside it,
at the 0.01 level of significance we can't reject the null hypothesis that the population
mean is 165.
Question 2
This is simply an exercise in using the standard normal table. Define X as the temperature
during June.
a) Find Prob( 21.11 < X < 26.66)
This the same as finding
Prob (X < 26.66) - Prob (X < 21.11)
So let's find each of them. Recall again that given a normally distributed variable, we can
"standarize" it by subtracting its mean dividing by its std dev. So we have:
Prob ( X < 26.66)
=Prob( (X-20)/3.33 < (26.66-20)/3.33)
=Prob( Z
< 2)
Now it's just a matter of checking a standard normal table. We get from it that Prob(Z<2)
= 0.9772
Now we just do the same with the other term:
Prob (X < 21.11)
=Prob ( (X-20)/3.33 < (21.11-20)/3.33 )
=Prob (
Z
< 0.33)
The standard normal table tells us that this probability is 0.6293. Therefore:
Prob( 21.11 < X < 26.66) = Prob (X < 26.66) - Prob (X < 21.11) = 0.9772 - 0.6293 =
0.3479
So P(A) = 0.3479.
b) The event A and B is that X is between 21.11 and 26.66 AND that it is higher than 25.
Therefore, event A and B is that X is between 25 and 26.66. The procedure for finding
the solution is basically the same as in question (a):
Prob ( 25 < X < 26.66)
After doing all the table checking and variable standarizing, we get that this probability is
0.0438.
[Please note that your result could differ very slightly due to some rounding]
c) The freezing point is 0 degrees, so we're asked to check:
Prob (X < 0)
After standarizing (subtracting mean, dividing by std dev), we get that this is the same as:
Prob(Z < -6)
The prob of Z being lower than -4 is already almost zero. Therefore, we conclude that
P(C) = 0.
d) Here we're asked to find the probability that the temperature is below 0 given that the
temperature is higher than 25. Clearly, if the temperature is higher than 25, it can't be
lower than 0. Therefore, P(C given B) = 0.
e) Here we're aksed the probability that the temperature is not between 21.11 and 26.66
(not A) and is lower than 25 (not B). This is equivalent to asking what is the probability
that the temperature is lower than 21.11. So we have:
P ( X < 21.11)
Using the same procedure as before, we get that this probability is 0.6305
f) Here, we must first find P(B).
P( X > 25)
This probability is 0.0666. This number tells us that we can expect 6.66% of June's days
to have a temperature higher than 25. Since June has 30 days, the number of days in
which we can expect event B is 0.0666*30 = 1.998. So we can expect approximately 2
days in June to have a temperature higher than 25.