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This work is licensed under the Creative commons Attribution-Non-commercial-Share Alike 2.5 South
Africa License. You are free to copy, communicate and adapt the work on condition that you attribute the
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Chapter
9
THE t- AND F-DISTRIBUTIONS
KEYWORDS: t-distribution, degrees of freedom, pooled sample variance, F -distribution
Population variance unknown . . .
So far, for confidence intervals for the mean, and for tests of hypotheses about means,
we have always had to make the very restrictive assumption that the population variance
(or standard deviation) was known. Until the early 1900’s, whenever the population
variance σ 2 was unknown, the sample variance s2 was substituted in its place. This
is reasonably satisfactory when the samples are “large enough”, and consequently the
sample variance lies reliably close to the population variance. We shall see in this
chapter that when the sample size exceeds about 30 then it is reasonable to make this
substitution. Problems arise with small samples, because the sample variance s2 is then
very variable and can be far removed from the population variance σ 2 , which we think
of as being the one “true” value.
At the turn of the century, W.S. Gosset, who worked for an Irish brewery, needed to
do statistical tests using small samples. This motivated him to tackle the mathematical
statistical problem of how to use s2 , the estimate of σ 2 , in these tests. He solved the
problem. Because of his professional connections, he published the theory he developed
under the pen-name “Student”. Gosset in 1908 published the actual distribution of
t=
X −µ
√
s/ n
X−µ
√ has the standard normal distribution. But
We know, from chapter 8, that Z = σ/
n
when σ , the single “true” value for the standard deviation in the population, is replaced
by s, the estimate of σ , this is no longer true, although, as the sample size n increases,
it rapidly becomes an excellent approximation. But for small samples it is far from the
truth. This is because the sample variance s2 (and also s) is itself a random variable
— it varies from sample to sample. We will discuss the sampling distribution of s2 in
the next chapter.
We know that the size of the sample influences the accuracy of our estimates. The
larger the sample the closer the estimate is likely to be to the true value. Student’s
t-distribution takes account of the size of the sample from which s is calculated.
The shape of the t-distribution is similar to that of the normal distribution. However, the shape of the distribution varies with the sample size. It is longer-
199
200
INTROSTAT
or heavier-tailed than the normal distribution when the sample size is small. As the
sample size increases, the t-distribution and normal distribution become progressively
closer, and, ultimately, they are identical. The standard normal distribution, and two
t-distributions are plotted.
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N (0, 1)
t
t
−3
−2
−1
0
1
2
3
Degrees of freedom . . .
In order to gain some insight into the notion of “degrees of freedom”, consider again
the definition of the sample variance:
n
X
1
s =
(xi − x̄)2 .
n−1
2
i=1
The terms xi − x̄ are the deviations of each of the xi from the sample mean. To
achieve a given sample mean for, say, six numbers, five of these can be chosen at will,
but the last is then fully determined. Suppose we are given the information that the
mean of six numbers is 5 and that the first five of the six numbers are 4, 9, 5, 7 and 3.
The sixth number must be 2, otherwise the sample mean would not be 5. It is fixed, it
has no “freedom”. In general if we are told that the mean of n numbers is x̄, and that
the first n − 1 numbers are x1 , x2 , . . . , xn−1 , then it is easy to see that xn must be
given by
xn = nx̄ − x1 − x2 · · · − xn−1 .
In other words, once we are given x̄ and the first n − 1 of the xi we have enough
information to compute the sample variance. Thus, only n − 1 of the deviation terms
xi − x̄ in s2 contain real information; the last term is just a formality, but it must be
included! We say that s2 , based on a sample of size n, has n − 1 degrees of freedom.
(This is part of the reason why the formula for the sample variance s2 calls for division
by n − 1, and not n.)
We will be encountering the concept of degrees of freedom regularly. We have a
simple rule which helps in making decisions about degrees of freedom.
THE DEGREES OF FREEDOM RULE
For each parameter we need to estimate prior to evaluating the current
parameter of interest, we lose one degree of freedom.
CHAPTER 9. THE t- AND F-DISTRIBUTIONS
201
For example, when we use s2 to estimate σ 2 , we first need to use x̄ to estimate µ.
So we lose one degree of freedom. There will be further examples later on in which we
will “lose” two or more degrees of freedom!
Because the shape of the t-distribution varies with the sample size, there is a whole
family of t-distributions. It is therefore necessary to have some means of indicating
which t-distribution is being used in a specific situation. We do this by using a subscript.
Intuition suggests that the subscript should be the sample size, but it turns out that the
sensible subscript is the degrees of freedom of the standard deviation, which is one less
than the sample size. So we use the notation
tn−1 =
X −µ
√
s/ n
and say that the expression on the right hand side has the t-distribution with n − 1
degrees of freedom, or simply the tn−1 -distribution.
As for the normal distribution, we need tables for looking up values for the tdistribution. The shapes of the t-distributions are dependent on the degree of freedom;
thus we cannot get away with a single table for all t-distributions as we did with the
normal distribution. We really do appear to need a separate table for each number of
degrees of freedom. But we take a short cut, and we only present a selection of key
values from each t-distribution in a single table (Table 2). If you think about it, you
can now begin to understand why we can do this; even for the normal distribution, we
repeatedly only use a handful of values; z (0.05) = 1.64, z (0.025) = 1.96, z (0.01) = 2.33
and z (0.005) = 2.58 are by far the most frequently used percentage points of the standard
normal distribution. In Table 2, there is one line for each t-distribution; on that line,
we present 11 percentage points.
As mentioned previously, exact p-values are usually computed using statistical software. For the z -test we could calculate these exactly since our tables gave us probabilities associated with z -scores. But due to the fact that our t-tables report critical
values rather than probabilities, p-values can no longer be computed exactly. To obtain approximate p-values: (1) identify the appropriate degrees of freedom, (2) highlight
the relevant line (corresponding to the df) in the t-tables, (3) identify where the test
statistic would lie along that line and (4) determine the approximate probability (range)
associated with the test statistic by looking at the probabilities corresponding with the
two values on either side of the test statistic.
To illustrate, let’s assume we observed a t-test statistic of 3.2156 with 12 degrees of
freedom. If we look in the line (on the t-table) corresponding to 12 degrees of freedom,
we see that our value lies between 3.055 and 3.428. Now the tail probabilities associated
with these two values respectively are 0.005 and 0.0025. This implies that the area in
the tail associated with the test statistic would lie somewhere between 0.0025 and 0.005.
So we have that 0.0025 < p-value < 0.005 in this instance (for a upper tail test)! For a
two-tailed test, these values need to be multiplied by 2 and the p-value would lie between
0.005 and 0.01.
Confidence intervals (using the sample variance) . . .
Example 1A: Ten direct flights to Johannesburg from Cape Town took, on average, 103
minutes. The sample standard deviation was 5 minutes. Determine a 95% confidence
interval for the true mean flight duration.
We have x̄ = 103, s = 5 and n = 10. We know that the statistic (X − µ)/ √s has
n
the t-distribution with 9 degrees of freedom, denoted by t9 .
202
INTROSTAT
In Table 2, we see that, for t9 , the points between which 95% of the distribution
lies are −2.262 and +2.262.
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t∼t
−3 −2
−2.262
−1
0
1
2
2.262
3
Because 2 12 % (or 0.025) of the t9 distribution lies to the right of 2.262 we write
(0.025)
t9
= 2.262
and speak of the “2 12 % point of the t9 distribution”.
We can write
Pr(−2.262 < t9 < 2.262) = 0.95.
But
X −µ
√ ∼ t9 .
s/ 10
Therefore
X −µ
Pr −2.262 < √ < 2.262 = 0.95.
s/ 10
Manipulation of the inequalities, as done in the same context in Chapter 8, yields
s
s
Pr X − 2.262 √ < µ < X + 2.262 √
= 0.95.
10
10
Substituting X = 103 and s = 5 yields
s
5
X − 2.262 √ = 103 − 2.262 √ = 99.4
10
10
for the lower limit of the confidence interval and
s
5
X + 2.262 √ = 103 + 2.262 √ = 106.6
10
10
for the upper limit. Thus a 95% confidence interval for the mean flying time is given by
(99.4, 106.6).
CONFIDENCE INTERVAL FOR µ, WHEN σ 2 IS
ESTIMATED BY s2
If we have a random sample of size n from a population with a normal
distribution, and the sample mean is x and the sample variance is s2 ,
then confidence intervals for µ are given by
s
s
∗
∗
X − tn−1 √ , X + tn−1 √
n
n
where the t∗ values are obtained from the t-tables. For 95% confidence intervals, use the column in the tables headed 0.025. For 99%
confidence intervals, use the column headed 0.005.
203
CHAPTER 9. THE t- AND F-DISTRIBUTIONS
Example 2B: A random sample of 25 loaves of bread had a mean mass of 696 g and
a standard deviation of 7 g. Calculate the 99% confidence interval for the mean.
We have x = 696, and s = 7 has 24 degrees of freedom. Because we want 99%
(0.005)
confidence intervals, we must look up t24
in the t-tables:
(0.005)
t24
= 2.797.
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t∼t
−3 −2
−2.797
−1
0
1
2
3
2.797
Thus the 99% confidence interval for the mean is
√
√
(696 − 2.797 × 7/ 25 , 696 + 2.797 × 7/ 25)
which reduces to (692.08, 699.92).
Example 3C: An estimate of the mean fuel consumption (litres/100 km) of a new
car is required. A sample of 18 drivers each drive the car under a variety of conditions
for 100 km, and the fuel consumption is measured for each of the drivers. The sample
mean was calculated to be x̄ = 6.73 litres/100 km. The sample standard deviation was
s = 0.35 litres/100 km. Find the 99% confidence interval for the population mean.
Testing whether the mean is a specified value
(population variance estimated from the sample) . . .
Example 4A: A poultry farmer is investigating ways of improving the profitability of
his operation. Using a standard diet, turkeys grow to a mean mass of 4.5 kg at age 4
months. A sample of 20 turkeys, which were given a special enriched diet, had an average
mass of 4.8 kg after 4 months. The sample standard deviation was 0.5 kg. Using the
5% significance level test whether the new diet is effectively increasing the mass of the
turkeys.
We follow the standard hypothesis testing procedure.
1. H0 : µ = 4.5. As usual, for our null hypothesis we assume no change has taken
place.
2. H1 : µ > 4.5. A one-sided alternative is appropriate, because an enriched diet
should not cause a loss of mass.
3. Significance level: 5%
4. The rejection region is found by reasoning as follows. Because the population
variance is unknown, we need to use the t-distribution. The degrees of freedom for
t will be 19, because s is based on a sample of 20 observations. We will thus reject
(0.05)
(0.05)
H0 if the “observed t-value” exceeds t19 . From the t-tables, t19
= 1.729.
204
INTROSTAT
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t∼t
−3
−2
−1
0
1
2
1.729
3
5. The formula for the test statistic is
t=
X −µ
√ .
s/ n
We substitute the appropriate values, and compute
t=
4.8 − 4.5
√ = 2.68.
0.5/ 20
6. Because 2.68 > 1.729 we reject H0 , and conclude that, at the 5% significance
level, we have established that the new enriched diet is effective.
Example 5B: The average life of 6 car batteries is 30 months with a standard deviation
of 4 months. The manufacturer claims an average life of 3 years for his batteries. We
suspect that he is exaggerating. Test his claim at the 5% significance level.
1. H0 : µ = 36 months .
2. H1 : µ < 36 months .
3. Significance level : 5%.
4. Degrees of freedom is 6 − 1 = 5. So we use the t5 -distribution. From the form
of the alternative hypothesis and the significance level, we will reject H0 if the
(0.05)
(0.05)
= −2.015.
. From our tables −t5
observed t-value is less than −t5
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t∼t
−3
−2 −1
−2.015
0
1
2
3
− µ yields
5. Substituting into the test statistic tn−1 = X √
s/ n
t5 =
30 − 36
√ = −3.67
4/ 6
6. This lies in the rejection region: we conclude that the true mean is significantly
less than 36 months.
Example 6C: A purchaser of bricks believes that their crushing strength is deteriorating. The mean crushing strength had previously been 400 kg, but a recent sample of 81
bricks yielded a mean crushing strength of 390 kg, with a standard deviation of 20 kg.
Test the purchaser’s belief at the 1% significance level.
CHAPTER 9. THE t- AND F-DISTRIBUTIONS
205
Example 7C: The specifications for a certain type of ball bearings stipulate a mean
diameter of 4.38 mm. The diameters of a sample of 12 ball bearings are measured and
the following summarized data computed:
Σxi = 53.18
Σx2i = 235.7403.
Is the sample consistent with the specifications?
Comparing two sample means: matched pairs or paired
samples . . .
We’ll introduce the idea of a paired t-test via an example.
Example 8A: Consider a medicine designed to reduce dizziness. We want to know
if it is effective. We have 10 patients who complain of dizziness and we examine the
reduction in dizzy spells for each patient when taking the medicine. In other words, we
ask how many dizzy spells each patient had per month before taking the medication and
again while on (after) the medication. Since a specific “before” score can be linked to a
specific “after” score, taking a difference in the dizzy spells is a sensible measure of the
reduction in dizzy spells. Note the data below, where the difference in dizzy spells have
been calculated.
1
2
3 4 5 6
7
8
9 10
Before (B): 19 18 9 8 7 12 16 22 19 18
After (A):
17 24 12 4 7 15 19 25 16 24
d=B−A
2 -6 -3 4 0 -3 -3 -3 3 -6
The two samples have effectively been reduced to one, by taking the difference in
scores. Therefore, testing if the two populations (“before” and “after”) have the same
mean, is equivalent to testing if the (population) mean of the difference scores is 0. So,
the two paired samples have been reduced to one sample on which we can perform a
one-sample t-test.
The relevant summary statistics obtained from the is d¯ = −1.5 and sd = 3.57.
Performing the hypothesis test assuming a 5% level of significance (using the six-point
plan), we have the following:
1. H0 : µd = 0
2. H1 : µd > 0
3. α = 0.05
4. The critical value is 1.833
5. If the number of dizzy spells have reduced while on the treatment, the before
score should be greater than the after score. And since differences were defined by,
d = B − A, we’d expect them to be positive. Note that the sample information
is not consistent with this hypothesis; hypotheses are determined by what you are
interested in finding out, NOT by the data!
t9 =
−1.5 − 0
√ = −1.33
3.57/ 10
Note that n corresponds to the number of observations, and since the test was
performed on the difference scores, we count 10 differences.
206
INTROSTAT
6. Since the test statistic −1.33 < 1.833 we fail to reject the null hypothesis and
conclude that the medication isn’t effective at reducing dizzy spells.
Note: It is assumed that the population of difference scores is normally distributed
with a mean of µd . Also, in the case above the p-value associated with the test statistic
would have been between 0.8 and 0.9 (clearly resulting in a fail to reject decision).
As already mentioned, in this example we appear to have two sets of data - the
“before” data and the “after” data. But were these really two separate samples? If
we had 10 randomly selected patients for “before” and then a different 10 randomly
selected patients for “after” then the answer would be yes and the samples would be
independent of one another. But, since dizziness varies from patient to patient, we
looked at the same 10 patients and took two readings from each individual. Thus our
two sets of data were dependent and to perform the test we used a single sample of
differences. These dependent measures are known as repeated measures.
Example 9C: A study was conducted to determine if the pace of music played in a
shopping centre influences the amount of time that customers spend in the shopping
centre. To test the hypothesis, a sample of 8 customers was observed on two different
days. On the one day, the customers did their shopping while slow music was playing.
On the same day the next week, they did their shopping while fast music was playing.
Other than the pace of the music, circumstances were similar. The following observations
were made:
Type of music
Slow music (A):
Fast music (B):
d=A−B
Time spent in shopping centre (minutes)
67 52 98 24 55 43 48 70
58 44 80 30 48 42 40 63
9
8
18
-6
7
1
8
7
The following sample means and standard deviations were observed:
ā = 57.125
sa = 21.86
b̄ = 50.625
sb = 15.74
d¯ = 6.5
sd = 6.87
Do the data above provide sufficient evidence that the pace of the music has an
influence on the amount of time that shoppers spend in a shopping centre?
Another way for getting dependent samples is when individuals are paired on some
criteria to reduce spurious variation in measurements. As a consequence of the latter,
the test we used is often called the paired t-test.
Example 10A: Twenty individuals were paired on their initial rate of reading. One of
each pair was randomly assigned to method I for speed reading and the other to method
II. After the courses the speed of reading was measured. Is there a difference in the
effectiveness of the two methods?
Pair
Method I:
Method II:
d = I − II
1
1114
1032
82
2
996
1148
-152
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9
996
1032
-36
10
894
1012
-118
207
CHAPTER 9. THE t- AND F-DISTRIBUTIONS
The relevant summary statistics are
d¯ = −40 sd = 71.6 and n = 10 pairs
Once again we assume that this is a random representative sample from the population of differences and that the differences in reading speed are normally distributed
with mean µd .
1. H0 : µd = 0
2. H1 : µd 6= 0
3. t9 =
−40−0
√
71.6/ 10
= −1.77
4. The p-value for this statistic lies between 0.1 and 0.2 and
5. Therefore we do not to reject the null hypothesis (p-value is too large) and conclude
that there isn’t a difference in reading speed between the two methods.
Confidence intervals . . .
As before, it is also possible to construct confidence intervals for these differences.
Following the same logic as before and using the fact that
d¯ − µd
√ ∼ tn−1 ,
sd / n
the 95% confidence interval can be constructed by:
d¯ − µd
0.025
0.025
√
Pr −tn−1 <
< tn−1 = 0.95
sd / n
So
sd
¯ 0.025 sd
Pr d¯ − t0.025
n−1 √ < µd < d + tn−1 √
n
n
= 0.95
Therefore the 95% confidence interval is:
0.025 sd
0.025 sd
¯
¯
d − tn−1 √ ; d + tn−1 √
n
n
Example 10A continued: To compute the 95% confidence interval:
71.6
71.6
−40 − 2.626 √ ; −40 + 2.626 √
10
10
Which is equivalent to
(−40 − 51.216; −40 + 51.216)
And this results in a 95% confidence interval for difference in speed reading methods
of -91.22 to 11.22. Note that zero lies in this interval.
208
INTROSTAT
Example 8A continued: The 95% confidence interval is:
3.57
3.57
−1.5 − 2.626 √ ; −1.5 + 2.626 √
10
10
Which is equivalent to
(−1.5 − 2.554; −1.5 + 2.554)
The corresponding 95% confidence interval for difference in dizzy spells per month
is -4.05 to 1.05. Note again that zero lies in this interval.
Comparing two independent sample means (variance estimated from the sample) . . .
When we have small samples from two populations and want to compare their means,
the procedure is a little more complex than you might have expected. In Chapter 8, the
test statistic for comparing two means was
Z=
X 1 − X 2 − (µ1 − µ2 )
r
.
σ12 σ22
n1 + n2
You would anticipate that the test statistic now might be
X 1 − X 2 − (µ1 − µ2 )
r
.
s21
s22
n1 + n2
But, unfortunately, mathematical statisticians can show that this quantity does not have
the t-distribution. In order to find a test statistic which does have the t-distribution,
an additional assumption needs to be made. This assumption is that the population
variances in the two populations from which the samples were drawn are equal. The
examples below highlight this difference between comparing the means of two populations
when the variances are known and when the variances are estimated from the samples.
Example 11A: Two varieties of wheat are being tested in a developing country. Twelve
test plots are given identical preparatory treatment. Six plots are sown with Variety 1
and the other six plots with Variety 2 in an experiment in which the crop scientists
hope to determine whether there is a significant difference between yields, using a 5%
significance level.
The results were:
Variety 1 :
Variety 2 :
1.5
1.6
1.9
1.8
1.2
2.0
1.4
1.8
2.3
2.3
and
1.3
tons per plot
tons per plot
One of the plots planted with Variety 2 was accidently given an extra dose of fertilizer,
so the result was discarded. The means and standard deviation are calculated. They
are
x̄1 = 1.60
s1 = 0.42
n1 = 6
x̄2 = 1.90
s2 = 0.27
n2 = 5.
We follow the standard hypothesis testing procedure.
209
CHAPTER 9. THE t- AND F-DISTRIBUTIONS
1. H0 : µ1 − µ2 = 0.
2. H1 : µ1 − µ2 6= 0
(a two-tailed test).
3. Significance level : 5%.
4. & 5. Before we can find out the rejection region we need to know the “degrees of freedom”. The procedure is rather different from that in the test when the population
variances were known. Instead of working with the individual variances σ12 and
σ22 we assume that both populations have the same variance and we pool
the two individual sample variances s21 and s22 to form a joint estimate of the
variance, s2 . This assumption of equal variances is required by the mathematical
theory underlying the t-distribution, into which we will not delve. Of course, we
do need to check this assumption of equal variances, which we shall discuss later
in the chapter (the F -distribution).
The general formula for the pooled variance is
s2 =
(n1 − 1)s21 + (n2 − 1)s22
n1 + n2 − 2
where s21 is based on a sample of size n1 and s22 is based on a sample of size n2 .
In the above example, n1 = 6 and n2 = 5. Therefore
5 × (0.42)2 + 4 × (0.27)2
6+5−2
= 0.13
s2 =
Therefore, the pooled standard deviation
√
s = 0.13 = 0.361.
How many “degrees of freedom” does s have? s21 has 5 and s22 has 4. Therefore s2
has 5+4 = 9 degrees of freedom. In general, s2 has (n1 −1)+(n2 −1) = n2 +n2 −2
degrees of freedom. We lose two degrees of freedom because we estimated the two
parameters µ1 (by X 1 ) and µ2 (by X 2 ) before estimating s2 .
Thus we use the t-distribution with 9 degrees of freedom, and because we have
(0.025)
a two-sided alternative and a 5% significance level we need the value of t9
,
which from the tables is 2.262. So we will reject H0 if the observed t9 -value is
less than −2.262 or greater than 2.262.
....................
......
.....
.....
....
....
...
...
...
.
.
...
..
9
.
...
.
..
...
.
.
...
..
.
.
...
..
.
.
.
....
...
....
...
....
....
...
...
.....
.....
.
.
......
.
.
...
......
.
.
.
.
.
.......
.....
.
.
.
........
.
.
.
.....
...........
.
.
.
.
.
.
.
.
.
.
.................
.................
t∼t
−3 −2
−2.262
−1
0
1
2
2.262
3
The formula for calculating the test statistic in this hypothesis-testing situation,
the so-called two-sample t-test, is
tn1 +n2 −2 =
X 1 − X 2 − (µ1 − µ2 )
r
1
1
+
s
n1 n2
210
INTROSTAT
where X 1 and X 2 are the sample means, the value of µ1 − µ2 is determined by
the null hypothesis, s is the pooled sample standard deviation, and n1 and n2 are
the sample sizes.
Substituting our data into this formula:
1.60 − 1.90 − 0
r
t6+5−2 =
.
1 1
0.361
+
6 5
Thus
t9 = −1.372.
6. Because −1.372 does not lie in the rejection region we conclude that the difference
between the varieties is not significant.
Example 12B: A marketing specialist considers two promotions in order to increase
sales of do-it-yourself hardware in a supermarket. During a trial period the promotions
are run on alternative days. In the first promotion, a free set of drill bits is given if
the customer purchases an electric drill. In the second, a substantial discount is given
on the drill. The marketing specialist is particularly interested in the average amount
spent on do-it-yourself hardware by customers who took advantage of the promotion. On
the basis of randomly selected samples, the following data were obtained of the amount
spent on items of do-it-yourself hardware.
Promotion
n
x
s
Free gift
8
R490
R104
Discount
9
R420
R92
Test, at the 5% level of significance, whether there is any difference in the effectiveness of the two promotions.
1. H0 : µ1 = µ2 .
2. H1 : µ1 6= µ2 .
3. Significance level : 5%.
4. Degrees of freedom : n1 + n2 − 2 = 15. From t-tables, if the observed t15 value
exceeds 2.131, we reject H0 .
........................
.....
....
....
....
....
...
...
...
.
.
...
15
..
.
...
.
..
...
.
.
...
..
.
.
...
..
.
.
.
.
....
..
....
...
...
....
....
....
.....
.....
.
.
......
.
.
....
......
.
.
.
.
.
.......
....
.
.
.
.
.........
.
.
.
......
.............
.
.
.
.
.
.
.
.
.
.
.
............
.............
t∼t
−3
−2 −1
−2.131
0
1
2
2.131
5. We need first to calculate s2 , the pooled variance:
(n1 − 1)s21 + (n2 − 1)s22
n1 + n2 − 2
7 × 1042 + 8 × 922
=
15
= 9561.60
s2 =
3
211
CHAPTER 9. THE t- AND F-DISTRIBUTIONS
and so the pooled standard deviation is
s = 97.78.
We now calculate the observed test statistic
tn1 +n2 −2 =
t15 =
X 1 − X 2 − (µ1 − µ2 )
q
s n11 + n12
490 − 420 − 0
q
= 1.47.
97.78 18 + 19
6. Because 1.47 < 2.131, we cannot reject H0 , and we conclude that we cannot detect
a difference between the effectiveness of the two promotions.
Example 13C: Two methods of assembling a new television component are under consideration by management. Because of more expensive machinery requirements, method
B will only be adopted if it is significantly shorter than method A by more than a minute.
In order to determine which method to adopt a skilled worker becomes proficient in both
methods, and is then timed with a stopwatch while assembling the component by both
methods. The following data were obtained:
Method A
Method B
x1 = 7.72 minutes
x2 = 6.21 minutes
s1 = 0.67 minutes
s2 = 0.51 minutes
n1 = 17
n2 = 25
What decision should be taken?
Testing variances for equality . . .
The t-tests for comparing means of two populations, as introduced in the previous
section, required us to assume that the two populations had equal variances.
This assumption can be tested easily. Using the statistical jargon we have developed,
what we need is a test of H0 : σ12 = σ22 against H1 : σ12 6= σ22 . (Situations do arise when
we need one-sided alternatives and the test we will develop can handle both one-sided
and two-sided alternatives.)
Because s21 and s22 are our estimates of σ12 and σ22 , our intuitive feeling is that
we would like to reject our null hypothesis when s21 and s22 are “too far apart”. This
might suggest the quantity s21 − s22 as our test statistic. However, we cannot easily
find the sampling distribution of s21 − s22 . It turns out that the test statistic which
is mathematically convenient to use is the ratio F = s21 /s22 . The sampling distribution
of the statistic F is known as the F -distribution, or Fisher distribution. Sir Ronald
Fisher was a British statistician who was one of the founding fathers of the discipline of
Statistics.
Because variances are by definition positive, the statistic F is always positive. When
H0 : σ12 = σ22 is true, we expect the sample variances to be nearly equal, so that F will
be close to one. When H0 is false, and the population variances are unequal, then F
will tend to be either large or small, where in this context small means close to zero.
Thus, we accept H0 for F -values close to one, and we reject H0 when the F -value is
too large or too small. The rejection region is obtained from F -tables, but the shape of
probability density function for a typical F -distribution is shown here.
212
INTROSTAT
0.6
f (x)
0.4
0.2
0.0
.......
... ...
.. ....
..
....
...
.
..
...
...
...
...
...
..
...
...
...
...
...
...
...
...
...
...
...
...
...
....
...
..
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
....
...
....
...
....
.....
...
......
...
......
.......
...
........
...
..........
............
...
..................
...
..................................
..............................................................................
....
.............................
F -distribution
0
2
4
x
6
8
The most striking feature of the probability density function of the F -distribution
is that it is not symmetric. It is positively skewed, having a long tail to the right.
The mode (the x-value associated with the maximum value of the probability density
function) is less than one, but the mean is greater than one, the long tail pulling the
mean to the right. The lack of symmetry makes it seem that we will need separate tables
for the upper and lower percentage points. However, by means of a simple trick (to be
explained later), we can get away without having tables for the lower percentage points
of the F -distribution.
Because the F -statistic is the ratio of two sample variances, it should come as no
surprise to you that there are two degrees of freedom numbers attached to F — the
degrees of freedom for the variance in the numerator, and the degrees of freedom for the
denominator variance. It would therefore appear that we need an encyclopaedia of tables
for the F -distribution! To avoid this, it is usual to only present the four most important
values for each F -distribution; the 5%, 2.5%, 1% and 0.5% points. The conventional
way of presenting F -tables is to have one table for each of these percentage points; in
this book Table 4.1 gives the 5% points, Table 4.2 the 2.5% points, Table 4.3 the 1%
points and Table 4.4 the 0.5% points. Within each table, the rows and columns are used
for the degrees of freedom in the denominator and numerator, respectively.
Example 14A: In example 11A, the sample standard deviations were s1 = 0.42 and
s2 = 0.27. Let us test at the 5% level to see if the assumption of equal variances was
reasonable.
1. H0 : σ12 = σ22 .
2. H1 : σ12 6= σ22 .
3. Significance level : 5%.
4. The rejection region. If s21 , based on a sample of size n1 (and therefore having
n1 − 1 degrees of freedom), is the numerator, and s22 (sample size n2 , degrees
of freedom n2 − 1) is the denominator, then we say that F = s21 /s22 has the
F -distribution with n1 − 1 and n2 − 1 degrees of freedom. We write
s21
∼ Fn1 −1,
s22
n2 −1 .
In example 11A, the sample sizes for s21 and s22 were 6 and 5 respectively. Thus
we use the F -distribution with 6 − 1 and 5 − 1 degrees of freedom, i.e. F5,4 .
213
CHAPTER 9. THE t- AND F-DISTRIBUTIONS
Because we have a two-sided test at the 5% level, we need the upper and lower 2 12 %
points of F5,4 . This means that we must use Table 4.2, and go to the intersection
of column 5 and row 4, where we find that the upper 2 12 % point of F5,4 is 9.36.
(0.025)
We write F5,4
= 9.36. (Notice that, in F -tables, the usual matrix convention
of putting rows first, then columns, is not adopted.)
0.6
f (x)
0.4
...
... ...
.. ...
... ....
...
..
.
..
..
...
..
...
..
...
...
...
...
...
...
..
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
5,4
...
...
...
...
...
...
...
...
...
..
...
...
...
...
...
...
....
...
....
...
....
...
.....
......
...
......
...
.......
........
...
(0.025)
..........
...
............
..................
...
5,4
................................
...
..................................................................
.........................................
..
F
0.2
F
0.0
0
4
8
= 9.36
12
x
We will reject H0 if our observed F value exceeds 9.36. The tables do not enable
us to find the lower rejection region, but for the reasons explained below, we do
not in fact need it.
5. The observed F -value is
F = s21 /s22 = 0.422 /0.272 = 2.42.
6. Because 2.42 < 9.36, we do not reject H0 . We conclude that the assumption of
equal variances is tenable, and that therefore it was justified to pool the variances
for the two-sample t-test in example 11A.
The trick that enables us never to need lower percentage points of the F -distribution is to adopt the convention of always putting the numerically larger variance into
the numerator — so that the calculated F -statistic is always larger than one — and
adjusting the degrees of freedom. Let s21 and s22 have n1 − 1 and n2 − 1 degrees of
freedom respectively. Then, if s21 > s22 , consider the ratio F = s21 /s22 which has the
Fn1 −1,n2 −1 -distribution. If s22 > s21 , use F = s22 /s21 with the Fn2 −1,n1 −1 -distribution.
This trick depends on the mathematical result that 1/Fn1 −1,n2 −1 = Fn2 −1,n1 −1 .
Example 15B: We have two machines that fill milk bottles. We accept that both
machines are putting, on average, one litre of milk into each bottle. We suspect, however,
that the second machine is considerably less consistent than the first, and that the volume
of milk that it delivers is more variable. We take a random sample of 15 bottles from
the first machine and 25 from the second and compute sample variances of 2.1 ml2 and
5.9 ml2 respectively. Are our suspicions correct? Test at the 5% significance level.
1. H0 : σ12 = σ22 .
2. H1 : σ12 < σ22 .
3. Significance level : 5% (and note that we now have a one-sided test).
214
INTROSTAT
4. & 5. Because s22 > s21 , we compute
F = s22 /s21 = 5.9/2.1 = 2.81.
(0.05)
This has the Fn2 −1,n1 −1 = F24,14 -distribution. From our tables, F24,14 = 2.35
1.0
f (x)
0.5
...........
... .....
...
...
...
..
.
...
...
...
.....
..
..
..
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
24,14
...
...
...
...
...
...
...
...
...
..
...
...
...
...
...
....
....
....
....
...
.....
(0.05)
......
...
......
...
.......
24,14
.
.........
.
...........
..
.
...............
..............................
...
..................................................................
.................
F
F
0.0
0
1
2
x
= 2.35
3
4
6. The observed F -value of 2.81 > 2.35 and therefore lies in the rejection region.
We reject H0 and conclude that the second bottle filler has a significantly larger
variance (variability) than the first.
Example 16C: Packing proteas for export is time consuming. A florist timed how long
it took each of 12 labourers to pack 20 boxes of proteas under normal conditions, and
then timed 10 labourers while they each packed 20 boxes of proteas with background
music. The average time to pack 20 boxes of proteas under normal conditions was 170
minutes with a standard deviation of 20 minutes, while the average time with background
music was 157 minutes, with a standard deviation of 25 minutes. At the 5% significance
level, test whether background music is effective in reducing packing time. Test also the
assumption of equal variances.
An approximate t-test when the variances cannot be assumed to be equal . . .
The F -test described above ought always to be applied before starting the t-test to
compare the means. It is conventional to use a 5% significance level for this test. If we
do not reject the null hypothesis of equal variances, we feel justified in computing the
pooled sample variance which the two-sample t-test requires. What happens, however,
if the F -test forces us to reject the assumption that the variances are equal? We resort
to an approximate t-test, which does not pool the variances, but which makes an
adjustment to the degrees of freedom.
The test statistic is
(X 1 − X 2 ) − (µ1 − µ2 )
s
.
t∗ =
s22
s21
+
n1 n2
This statistic has approximately the t-distribution. By experimentation, researchers
have determined that the degrees of freedom for t∗ can be approximated by
. (s2 /n )2 (s2 /n )2 2
1
1
∗
2
2
2
+ 2
n = (s1 /n1 + s2 /n2 )
− 2.
n1 + 1
n2 + 1
215
CHAPTER 9. THE t- AND F-DISTRIBUTIONS
This messy formula inevitably gives a value for n∗ which is not an integer. It is
feasible to interpolate in the t-tables, but we will simply take n∗ to be the nearest
integer value to that given by the formula above.
Example 17B: Personnel consultants are interested in establishing whether there is
any difference in the mean age of the senior managers of two large corporations. The
following data gives the ages, to the nearest year, of a random sample of 10 senior
managers, sampled from each corporations:
Corporation 1
Corporation 2
52
44
50
45
53
39
42
49
57
43
43
49
52
45
44
47
51
42
34
46
Conduct the necessary test.
We compute
x̄1 = 47.80
s1 = 6.86
x̄2 = 45.00
s2 = 3.20
We first test for equality of variances:
1. H0 : σ12 = σ22 .
2. H1 : σ12 6= σ22 .
3. F = s22 /s22 = 6.862 /3.202 = 4.60.
4. Using the F9,9 -distribution, and remembering that the test is two-sided, we see
(0.025)
= 4.03), although not
that this F -value is significant at the 5% level (F9,9
(0.005)
significant at the 1% level (F9,9
= 6.54).
5. The conclusion is that the population variances are not equal (F9,9 = 4.60, P <
0.05).
Thus pooling the sample variances is not justified, and we have to use the approximate t-test:
1. H0 : µ1 = µ2 .
2. H1 : µ1 6= µ2 .
3. Substituting into the formula for the approximate test statistic yields
(47.8 − 45.0)
= 1.17.
t∗ = r
6.862
3.202
+
10
10
Substituting into the degrees of freedom formula yields
(
2 . )
(6.862 /10)2
3.202
(3.202 /10)2
6.862
∗
−2
+
+
n =
10
10
11
11
= 13.57 ≈ 14
(0.100)
4. Because t14
level.
= 1.345, we accept the null hypothesis even at the 20% significance
216
INTROSTAT
5. We conclude that there is no difference in the mean age of senior employees between
the two corporations (t14 = 1.17, P > 0.20).
Example 18C: A particular business school requires a satisfactory GMAT examination
score as its entrance requirement. The admissions officer believes that, on average, engineers have higher GMAT scores than applicants with an arts background. The following
GMAT scores were extracted from a random sample of applicants with engineering and
arts backgrounds.
Engineering
Art
600
550
650
450
640
700
720
420
700
750
620
500
740
520
650
Investigate the admission officer’s belief.
Example 19C: The dividend yield of a share is the dividend paid by the share during
a year divided by the price of the share. A financial analyst wants to compare the
dividend yield of gold shares with that of industrial shares listed on the Johannesburg
Stock Exchange. She takes a sample of gold shares and a sample of industrial shares
and computes the dividend yields. What conclusions did she come to?
Gold
shares
Industrial
shares
3.6
4.6
3.2
2.5
4.0
4.0
2.5
8.4
3.9
3.2
8.4
5.2
5.0
15.5
8.7
4.7
2.7
5.7
2.7
3.1
3.7
3.6
3.1
5.5
4.6
4.1
5.3
6.5
3.5
6.2
4.3
3.6
4.5
3.5
3.9
3.7
5.6
4.5
4.0
1.6
5.1
3.1
6.8
The data are summarized as:
n
x
s
Gold
Industrial
20
4.675
2.677
23
4.710
2.004
A important footnote to t-tests and F -tests . . .
Whenever the t-test or F -test is applied, it is assumed that the population from
which the sample was taken has a normal distribution. Can we test this assumption?
And how should we proceed if the test shows that the distribution of the data is not a
normal distribution?
The first of these questions can be answered using the methods of chapter 10. The
answer to the second question is that methods for doing tests for non-normal data do
exist, but are beyond the scope of this book. For the record, they are called nonparametric tests — the tests using the normal distribution and the t-distribution are
known as parametric tests.
Another question. What should we do when there are three (or more) populations
which we want to compare simultaneously? In example 18C, we might have wanted to
do a comparison between students with engineering, arts and science backgrounds, and
217
CHAPTER 9. THE t- AND F-DISTRIBUTIONS
START
Is the population
variance known or
estimated?
Population variance
known
Population variance
unknown
Is there one
sample or two?
Is there one
sample or two?
Two samples.
Test statistic:
One sample.
Test statistic:
X̄ − µ
√
Z=
σ/ n
∼ N (0, 1)
Z=
One sample.
Test statistic:
X̄1 − X̄2 − (µ1 − µ2 )
s
σ12 σ22
+
n1 n2
X̄ − µ
√
s/ n
∼ tn−1
t=
∼ N (0, 1)
All t-tests assume that
the samples come from
normal distributions
Two samples.
First test the assumption that
variances are equal
Test statistic:
F =
s21
s22
or
F =
s22
s21
∼ Fn1 −1,n2 −1 or ∼ Fn2 −1,n1 −1
Use 5% significance level, F > 1
Is assumption accepted or rejected?
Assumption rejected.
Test statistic:
Assumption accepted.
Test statistic:
X̄1 − X̄2 − (µ1 − µ2 )
r
1
1
s
+
n1 n2
∼ tn1 +n2 −2
t⋆ =
t=
where
s=
s
(n1 − 1)s21 + (n2 − 1)s22
n1 + n2 − 2
X̄1 − X̄2 − (µ1 − µ2 )
s
s21
s2
+
n1 n2
∼ tn⋆
where
2
s22
s21
+
n1 n2
⋆
n =
2
2 − 2
s21 /n1
s22 /n2
+
n1 + 1
n2 + 1
Figure 9.1: Decision Tree for Hypothesis Testing on Means
218
INTROSTAT
to test the null hypothesis that there is no difference between the mean GMAT scores
for these three groups of students. The method to use then is called the analysis of
variance, usually abbreviated to ANOVA.
A summary of hypothesis testing on means . . .
The decision tree displayed in Figure 9.1 and the comments below aim to give
clear guidelines to help you decide which test to apply, and presents the formulae for all
the test statistics of Chapters 8 and 9.
If the sample is large, greater than 30 say, then the central limit theorem applies
and X has a normal distribution. If the sample is smaller than 30, then X will have
a normal distribution if the population from which the sample is drawn has a normal
distribution. If the sample is small, and the underlying population does not have a
normal distribution, these techniques are not valid, and “non-parametric” statistical
tests must be applied. These tests fall beyond the scope of our course.
If the sample variance is estimated from a large sample, then it is common practice
to assume that σ 2 = s2 and to use the normal distribution (i.e. the methods of Chapter
8, where the population variance was assumed known). In any case, the t-distribution
is nearly identical to the normal distribution for large degrees of freedom, and the
percentage points are almost equal. We will adopt the convention that for degrees of
freedom 30 or fewer the t-distribution is used, between 30 and 100 use of both the tand the normal distribution is acceptable, and over 100 the normal distribution is mostly
used. This note only applies if the variance is estimated.
Example 20B: Use the decision tree to decide which test to apply. An experiment
compared the abrasive wear of two different laminated materials. Twelve pieces of
material 1 and 10 pieces of material 2 were tested and in each case the depth of wear
was measured. The results were as follows:
Material 1
Material 2
x1 = 8.5 mm
x2 = 8.1 mm
s1 = 4 mm
s2 = 5 mm
n1 = 12
n2 = 10
Test the hypothesis that the two types of material exhibit the same mean abrasive
wear at the 1% significance level.
Begin at START
The population variance is estimated. Go right. There are samples from two
populations. Go right again. The assumption that the variances are equal is
accepted. Check this for yourself. Go left.
Pool the variances and use the test statistic with n1 + n2 − 2 degrees of freedom.
Do this example as an exercise.
Example 21C: In an assembly process, it is known from past records that it takes
an average of 3.7 hours with a standard deviation of 0.3 hours to assemble a certain
computer component. A new procedure is adopted. The first 100 items assembled using
the new procedure took, on average, 3.5 hours each. Assuming that the new procedure
did not alter the standard deviation, test whether the new procedure is effective in
reducing assembly time.
219
CHAPTER 9. THE t- AND F-DISTRIBUTIONS
Solutions to examples
...
√
3C 6.73 ± 2.898 × 0.35/ 18, which is (6.49, 6.97)
6C t80 = −4.50 < −2.374, reject H0 .
7C t11 = 2.34, P < 0.05, significant difference.
9C t7 = 2.68, P < 0.05, significant difference.
13C t40 = 2.80, P < 0.005, significant. Adopt new method.
16C F9,11 = 1.56 < 3.59, cannot reject H0 and pool variances. t20 = 1.36 < 1.725,
cannot reject H0 .
(0.025)
18C F = 6.28 > F6,7
= 5.12. Therefore cannot pool variances. t⋆ = 2.18, n⋆ = 8.2,
so use degrees of freedom 8. Using one-sided test, there is a significant difference
(P < 0.05).
19C F19,22 = 1.784, so variances can be pooled. Two-sample t-test yields t41 =
−0.0244, no significant difference. Dividend yield on gold shares not significantly
different from that on industrial shares (t41 = −0.0244, P > 0.20).
21C Path through flow chart: start, standard deviation
known, go left, one sample, go
left again, and use test statistic z = (X − µ) √σn ∼ N (0, 1).
Exercises on confidence intervals
...
9.1 Find the 95% confidence interval for the mean salary of teachers if a random sample
of 16 teachers had a mean salary of R12 125 with a standard deviation of R1005.
∗ 9.2
We want to estimate the mean number of items of advertising matter received
by medical practitioners through the post per week. For a random sample of 25
doctors, the sample mean is 28.1 and the sample standard deviation is 8. Find
the 95% confidence limits for the mean.
9.3 Over the past 12 months the average demand for sulphuric acid from the stores
of a large chemical factory has been 206 ℓ; the sample standard deviation has
been 50 ℓ. Find a 99% confidence interval for the true mean monthly demand for
sulphuric acid.
∗ 9.4
A sample of 10 measurements of the diameter of a brass sphere gave mean x = 4.38
cm and standard deviation s = 0.06 cm. Calculate (a) 95% and (b) 99% confidence
intervals for the actual diameter of the sphere. Why is the second confidence
interval longer than the first?
220
INTROSTAT
Exercises on hypothesis testing (one sample) . . .
∗ 9.5
In a textile manufacturing process, the average time taken is 6.4 hours. An innovation which, it is hoped, will streamline the process and reduce the time, is
introduced. A series of 8 trials used the modified process and produced the following results:
6.1
5.9
6.3
6.5
6.2
6.0
6.4
6.2.
Using 5% significance level, decide whether the innovation has succeeded in reducing average process time.
9.6 In 1989, the Johannesburg Stock Exchange (JSE) boomed, and the Allshare Index
showed an annual return of 55.5%. A sample of industrial shares yielded the
following returns:
54.5
52.8
47.8
56.1
42.3
23.2
59.8
52.5
33.1
65.7
49.7
47.5
16.0
32.5
50.3
46.7
Test, at the 5% significance level, whether the performance of industrial shares
lagged behind the market as a whole.
9.7 The mean score on a standardized psychology test is supposed to be 50. Believing
that a group of psychologists will score higher (because they can “see through” the
questions), we test a random sample of 11 psychologists. Their mean score is 55
and the standard deviation is 3. What conclusions can be made?
∗ 9.8
The specification quoted by ABC Alloys for a particular metal alloy was a melting
point of 1660 ◦ C. Fifteen samples of the alloy, selected at random, had a mean
melting point of 1648 ◦ C with a standard deviation of 45 ◦ C. Is the melting point
lower than specified?
Exercises using F -test for equality of variances . . .
9.9 If independent random samples of size 10 from two normal populations have sample
variances s21 = 12.8 and s22 = 3.2, what can you conclude about a claim that the
two populations have the same variance? Use a 5% significance level.
∗ 9.10
From a sample of size 13 the estimate of the standard deviation of a population
was calculated as 4.47 and from a sample of size 16 from another population the
standard deviation was 8.32. Can these populations be considered as having equal
variances? Use a 5% level of significance.
221
CHAPTER 9. THE t- AND F-DISTRIBUTIONS
Exercises on hypothesis testing (two samples) . . .
9.11 The national electricity supplier claims that switching off the hot water cylinder
at night does not result in a saving of electricity. In order to test this claim
a newspaper reporter obtains the co-operation of 16 house owners with similar
houses and salaries. Eight of the selected owners switch their cylinders off at
night. The consumption of electricity in each house over a period of 30 days is
measured; the units are kWh (kilowatt-hours). The following data are collected:
OFF GROUP
n1 = 8
x̄1 = 680
s21 = 450
ON GROUP
n2 = 8
x2 = 700
s22 = 300
Test, at the 5% level, whether there is a significant saving in electricity if the
cylinder is switched off at night. Test the assumption that the variances are equal,
also at the 5% level.
∗ 9.12
A comparison is made between two brands of toothpaste to compare their effectiveness at preventing cavities. 25 children use Hole-in-None and 30 children use
Fantoothtic in an impartial test. The results are as follows:
Sample size
Average number of new cavities
Standard deviation
24
1.6
0.7
30
2.7
0.9
At the 1% level of significance, investigate whether one brand is better than the
other.
9.13 A company claims that its light bulbs are superior to those of a competitor on
the basis of a study which showed that a sample of 40 of its bulbs had an average
lifetime of 522 hours with a standard deviation of 28 hours, while a sample of
30 bulbs made by the competitor had an average lifetime of 513 hours with a
standard deviation of 24 hours. Test the null hypothesis µ1 − µ2 = 0 against a
suitable one-sided alternative to see if the claim is justified.
∗ 9.14
The densities of sulphuric acid in two containers were measured, four determinations being made on one and six on the other. The results were:
(1) 1.842
1.846
1.843
1.843
(2) 1.848
1.843
1.846
1.847
1.847
1.845
Do the densities differ at the 5% significance level,
(a) if there is no reason beforehand to believe that there is any difference in
density between the containers?
(b) if we have good reason to suspect that, if there is any difference, the first
container will be less dense than the second?
9.15 A teacher used different teaching methods in two similar statistics classes of 35
students each. Each class then wrote the same examination. In one class, the
mean was x1 = 82% with s1 = 3%. In the other class, the results were x2 = 77%
and s2 = 7%. Test to see if this provides the teacher with evidence that one
teaching method is superior to the other. Use a 5% significance level.
222
INTROSTAT
9.16 Two drivers, A and B, do fuel consumption tests on a single car. The cars are refuelled every 100 km, and the number of litres required to refill the tank measured.
Drivers A and B drive 2100 km and 2800 km, respectively, and so are refuelled
21 and 28 times. Driver B has recently read a pamphlet entitled Fuel Economy
Tips, and has been putting these ideas into practice. The results are summarized
in the table below:
Driver
Sample size
Sample mean
(ℓ/100 km)
Sample standard deviation
(ℓ/100 km)
A
B
21
28
9.03
8.57
1.73
0.89
Is Driver B more economical than Driver A?
Exercises using normal, t- and F -distributions . . .
9.17 The following statistics were calculated from random samples of daily sales figures
for two departments of a large store:
Department
Sample size
Mean daily sales (rands)
Standard deviation (rands)
Hardware
Crockery
15
1400
180
15
1250
120
The sales manager feels that mean sales in hardware is significantly higher than in
crockery. Test this idea statistically using a 1% significance level.
∗ 9.18
Travel times by road between two towns are normally distributed; a random
sample of 16 observations had a mean of 30 minutes and a standard deviation of
5 minutes.
(a) Find a 99% confidence interval for the mean travel time.
(b) Estimate how large a sample would be needed to be 95% sure that the sample
mean was within half a minute of the population mean. You need to modify
the formula for estimating sample sizes in Chapter 8 to take account of the
fact that you are given a sample standard deviation based on a small sample
rather than the population standard deviation.
9.19 An insurance company has found that the number of claims made per year on a
certain type of policy obeys a Poisson distribution. Until five years ago, the rate
of claims averaged 13.1 per year. New restrictions on the acceptance of this type
of insurance were introduced five years ago, and since then 51 claims have been
made.
(a) Test whether the restrictions have been effective in reducing the number of
claims.
(b) Find an approximate 95% confidence interval for the average claim rate under
the new restrictions.
223
CHAPTER 9. THE t- AND F-DISTRIBUTIONS
9.20 A new type of battery is claimed to have two hours more life than the standard
type. Random samples of new and standard batteries are tested, with the following
summarized results:
New
Standard
Sample size
Sample mean
(hours)
Sample standard
deviation (hours)
94
42
39.3
36.8
7.2
6.4
Test the claim at the 5% significance level.
∗ 9.21
The mean commission of floor-wax salesmen has been R6000 per month in the
past. New brands of wax are now providing stiffer competition for the salesmen,
but inflation has pushed up the commission per sale. Management wishes to test
whether the figure of R6000 per month still prevails, and examines a sample of 120
recent monthly commission figures. They are found to have a mean of R5850 and
a standard deviation of R800. Test at a 5% significance level whether the mean
commission rate has changed.
9.22 Two bus drivers, M and N, travel the same route. Over a number of journeys the
times taken by each driver to travel from bus stop 5 to bus stop 19 were noted.
The summarized results are presented below:
Driver
Trips
Mean
(minutes)
Standard Deviation
(minutes)
M
N
12
21
18.1
21.3
1.9
3.9
(a) Test whether driver M is more consistent in journey times than driver N.
(b) Test whether there is a significant difference in the times taken by each driver.
9.23 It is necessary to compare the precision of two brands of detectors for measuring
mercury concentration in the air. The brand B detector is thought to be more
accurate than the brand A detector. Seven measurements are made with a brand
A instrument, and six with a brand B instrument one lunch hour. The results
(micrograms per cubic metre) are summarized as follows:
x̄A = 0.87
sA = 0.019
x̄B = 0.91
sB = 0.008
At the 5% significance level, do the data provide evidence that brand B measures
more precisely than brand A?
9.24 Show that if we test at the 5% significance level (using either the t- or normal
distributions) the null hypothesis H0 : µ = µ0 against H1 : µ 6= µ0 , we will reject
H0 if and only if µ0 lies outside the 95% confidence interval for µ.
224
∗ 9.25
INTROSTAT
The health department wishes to determine if the mean bacteria count per ml
of water at Zeekoeivlei exceeds the safety level of 200 per ml. Ten 1 mℓ water
samples are collected. The bacteria counts are:
225
210
185
202
216
193
190
207
204
220
Do these data give cause for concern?
∗ 9.26
A normally distributed random variable has standard deviation σ = 5. A sample
was drawn, and the 95% confidence interval for the mean was calculated to be
(70.467, 73.733). The experimenter subsequently lost the original data. Tell him
(a) what the sample mean of the original data was, and
(b) what size sample he drew.
9.27 The following are observations made on a normal distribution with mean µ and
variance σ 2 :
50
53
47
51 49
(a) Find 95% and 99% confidence intervals for µ
(i) assuming σ 2 = 5
(ii) assuming σ 2 is unknown, and has to be estimated by s2 .
(b) Comment on the relative lengths of the confidence intervals found in (a).
Solutions to exercises
...
9.1 (11 590, 12 660)
9.2 (24.80, 31.40)
9.3 (161, 251)
9.4 (a) (4.335, 4.425)
(b) (4.315, 4.445)
The higher the level of confidence, the wider the interval.
9.5 t7 = −2.83 < −1.895, reject H0 .
9.6 t15 = −2.97 < −1.753, reject H0 , industrial shares performed worse than the
All-Share Index.
9.7 t10 = 5.53, P < 0.0005, very highly significant.
9.8 t14 = −1.03, P > 0.10, at this stage cannot reject H0 , the evidence is consistent
with the specifications, but further investigation is warranted.
9.9 F9,9 = 4.0 < 4.03, cannot reject H0 .
9.10 F15,12 = 3.46 > 3.18, reject H0 .
9.11 F7,7 = 1.50 < 4.99, cannot reject H0 and thus pooling of variances justified.
t14 = −2.066 < −1.761, reject H0 .
CHAPTER 9. THE t- AND F-DISTRIBUTIONS
225
9.12 F29,24 = 1.65 < 2.22, cannot reject H0 , pool variances.
t53 = −4.98 < −2.672, reject H0 .
9.13 F39,29 = 1.36 < (approx) 1.79, cannot reject H0 and pool variances.
t68 = 1.41, 0.05 < P < 0.10, insignificant.
9.14 F3,5 = 1.07 < 7.76, cannot reject H0 and pool variances.
(a) t8 = 2.19 < 2.306, cannot reject H0 .
(b) t8 = 2.19 > 1.860, reject H0 .
9.15 F34,34 = 5.44 > (approx) 2.30, reject H0 at 1% level. Variances cannot be pooled.
Degrees of freedom
n∗ = 46.8 ≈ 47. t∗ = 3.88 > 2.01,
reject H0 .
9.16 F20,27 = 3.78, P < 0.01, significant, so variances cannot be pooled. Degrees of
freedom
n∗ = 28.7 ≈ 29. t∗ = 1.11, P < 0 > 20,
insignificant.
9.17 F14,14 = 2.25 < 2.48, cannot reject H0 and hence pool the variances. t28 = 2.69 >
2.467, reject H0 .
9.18 (a) (26.32 , 33.68)
(b) n = (t⋆m−1 s/L)2 = (2.131 × 5/ 12 )2 = 455, where m is the size of the sample
used for estimating s (in this case 16), so the that degrees of freedom for t
is 15.
9.19 (a) z = −1, 79, P < 0.05, significant reduction.
(b) (7.4 , 13.0).
9.20 F93,41 = 1.27 < (approx) 1.59, cannot reject H0 and pool variances (s = 6.965).
t134 = 0386 < 1.656, cannot reject H0 .
9.21 t119 = −2.05 < −1.98, reject H0 .
9.22 (a) F20,11 = 4.214, P < 0.05, significant differences in variances. Pooling not
justifiable.
(b) n∗ = 32, t∗ = −3.16, P < 0.005, significant difference in means.
9.23 F6,5 = 5.64 > 4.95, reject H0 .
9.25 t9 = 1.25, P < 0.20, insignificant.
9.26 (a) x̄ = 72.1
(b) n = 36.
9.27 (a) The confidence intervals are:
95%
99%
(i)
(48.04 , 51.96)
(47.42 , 52.48)
(ii)
(47.22 , 52.78)
(45.40 , 54.60)
(b) 99% confidence intervals are wider than 95% confidence intervals. If σ 2 is
estimated by s2 , the confidence interval is wider.
226
INTROSTAT

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