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Series Editor: Patrick Johnson
MATHEMATICS WORKSHEETS FOR SCHOOLS
Volume 1: Number 2
a.
b.
Curriculum Links:
i. Hypothesis Testing
Levels:
i. Leaving Certificate Mathematics
ii. Junior Certificate Science
Maths ensures quality in
the Bio-medical Industry
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Mathematical Information
What is a hypothesis test?
A hypothesis test examines a claim about some characteristic of a
population, e.g. average length of nails produced by a company.
The population is very large, generally too large to test, and so
we draw a random sample from the population. This sample
should be large enough (unbiased) so that it doesn’t inadvertently
favour the characteristic that we are testing. Additionally, by
insuring that the sample is large enough means that it will be
representative of the population. Every sample is different, so the
sample characteristic value that we are testing (test statistic) varies
from sample to sample, i.e., it has a range. Therefore we set a
range of values that we believe covers the range for the test
statistic, and if the population value is within this range, we agree
with the claim being tested.
The one-sample Z-test is one of many procedures available for
hypothesis testing. The one-sample Z-test allows us to determine
whether the difference between the sample characteristic mean
value (x) and the population mean (µ) is statistically significant.
For large sample sizes many distributions are approximately
normally distributed and so the Z-test is commonly used when
testing hypotheses. The normal distribution (also referred to as
the Gaussian distribution) is a continuous probability distribution
that is often used as an approximation to describe real-valued data
that tend to cluster around a single mean value. The graph of the
A coronary stent is a metal tube used to unblock the arteries whose insertion
can help prevent a heart attack.
normal distribution is symmetrical, bell-shaped, centered about
its mean, with its spread determined by its standard deviation (see
Figure 1 overleaf). Examples of data that can be represented by
the normal distribution include height, weight, IQ (Intelligence
Quotient) and most manufacturing processes.
Hypothesis testing is an example of statistical inference, which
refers to using information from a sample to draw a conclusion
about the population. We use a hypothesis test to make inferences
about one or more populations when sample data are available.
NCE-MSTL and Engineers Ireland Mathematics Worksheets for Schools — Vol. 1 Number 2.
Series Ed. Patrick Johnson; Writing Team Patrick Johnson, Tim Brophy and Barry Fitzgerald.
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A coronary stent is a metal tube used to unblock the arteries
whose insertion can help prevent a heart attack. A stent is usually
inserted into an artery by key-hole surgery, a surgical procedure
which is far less stressful for a patient than open-heart by-pass
surgery. The stent is inserted using a device called a balloon
catheter. Once in position, the catheter is inflated and opens the
stent compressing the cholesterol that has built up in the artery.
This results in a widening of the passage through which blood
flows in the artery.
Figure 1:The standard normal distribution.
How does a hypothesis test work?
There are two hypotheses in a significance test—the null and
alternative hypotheses. A null hypothesis (H0) is the hypothesis
to be tested. It is assumed to be true unless the data indicates
otherwise. The null hypothesis usually assumes any variation in
the data is due to chance, whereas the alternative hypothesis
(Ha) assumes any variation is a real effect. The alternative
hypothesis (Ha) is the opposite to the null hypothesis.
All hypothesis tests follow the same steps:
1. Assume H0 is true.
2. Determine how different the sample is from what you
expected under the assumption of H0 being true.
3. If the test statistic is sufficiently unlikely under the assumption
that H0 is true, then reject H0 in favour of Ha.
Once we reach a conclusion in terms of H0 (e.g. Reject H0 or Fail
to reject H0) the final step is to interpret this conclusion in terms
of the original problem statement i.e. write our conclusion down
in simple English for everyone to understand.
Biological Information
The heart is a muscle that pumps blood containing food and
oxygen through blood vessels to all parts of the body. Like our
body, the heart also needs food and oxygen which reach it
through the coronary arteries. However, if a substance known as
cholesterol sticks to the inner walls of the arteries, they become
narrow which leads to a decrease in oxygen supply to the heart
muscle. This condition is known as coronary artery disease
(CAD) and can cause heart attacks.
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Example
Company X manufacturers coronary stents. They currently have
several suppliers of tubing but want to reduce this so that they
only have two main suppliers. Before deciding on which two
suppliers to keep they want to assess the performance of each
supplier. Once tubing is received in, it is processed through laser
machines which form the stent pattern. The company requires
that all stents weigh on average 161 mg(milligrams) with a
standard deviation (σ) of 5.2 mg1. Company X completed all
Monday’s production using Supplier A. They take a sample of
30 stents from the day’s production, weigh them and then
conduct an analysis to compare this to the required company
standard. Below is the measurements that were recorded for the
30 stents (n = 30).
166.0 168.9 164.5 161.0 169.0 160.5 161.7 157.8 158.3 163.3
166.5 159.0 154.6 167.7 167.1 164.0 152.7 161.9 157.6 164.1
165.5 160.9 170.0 153.8 161.7 175.6 157.5 165.5 157.2 158.0
An example of a hypothesis test on the mean of a single
sample where the population standard deviation is known is
now presented. The table below list all the information that
we have relating to the problem;
1
Population
Sample
Mean
µ = 161
x = calculated from sample
Standard Deviation
σ = known
s = calculated from sample
Size
N = unknown
n = 30
If the population standard deviation (σ) is known then we use this
value, otherwise we use the sample standard deviation (s) when
calculating the test statistic.
STEP 1
The first step is always to identify the hypothesis. This
means determining the null hypothesis (H0) and alternative
hypothesis (Ha) for the given question. Since we are only
interested in whether the mean weight of our sample varies from
the required mean weight this is a two-tailed test (sample mean
could be under or over the population mean). Therefore our
hypothesis is:
H0 : µ = 161
Ha : µ ≠ 161
STEP 2
The second step is to determine the test statistic. This is a Z-value,
determined from our sample, that we will be testing. The formula
used for determining the test statistic is:
TS =
significance level of the test. The significance level (α) of a
statistical hypothesis test is a fixed probability of wrongly
rejecting the null hypothesis H0, if it is in fact true. Usually, the
significance level is chosen to be 0.05 i.e. there is a 5% chance
of wrongly rejecting the null hypothesis. This means that out of
every 20 tests we are willing to accept that one will return an
incorrect result. This is a compromise between safety and
practicality. The critical value is a Z value and is often represented
as Z α/2. To determine the critical value divide the significance
level (0.05) by 2 since we have a two-tailed test. Find the cell in
the body of the normal tables (remember that it is the body of
the normal tables that we look up because the significant level is
a probability) with a value closest to 0.025 and read off the
associated Z value. Based on this Z value we can determine the
critical region (or rejection region) for our test. Calculate the
critical values for this test.
x–µ
StErr(x)
σ
where StErr(x) is the standard error of the sample and is given by
.
√n
Determine the test statistic for the sample in question.
STEP 5
We now ask the question:- “Does the test statistic lie in the critical
region?”
STEP 3
We now determine the distribution of our sample. Since n is
large, (≥30), we can assume that our data is approximately
normal and so this means that we will use the normal tables
when determining the critical value(s).
STEP 4
The critical value(s) for a hypothesis test is a threshold to which
the value of the test statistic (calculated in Step 2) in a sample is
compared to determine whether or not the null hypothesis is
rejected. Since this is a two-tailed test (determined in Step 1) it
means that there will be two critical values. The critical values are
found by looking up the normal tables once we know the
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STEP 6
Draw your conclusion from the result in Step 5. At this stage we
attempt to interpret the result in Step 5 and phrase it in terms of
the original problem specification.
Teacher Page
Solutions
STEP 2 SOLUTION
STEP 5 SOLUTION
To determine the test statistic we use the formula:
Since TS = 1.47 and the CR > 1.96, we can clearly see that the
test statistic does not lie in the critical, or rejection region.
TS =
x–µ
σ/√ n
STEP 6 SOLUTION
We first need to calculate x, the mean of the sample in
question. Summing all the values in the sample together and
dividing by 30 we find that x = 162.397.
Therefore the test statistic is:
TS
=
x–µ
σ/√ n
=
162.397 – 161
≈ 1.47
5.2 / √ 30
STEP 4 SOLUTION
Comments and Suggestions
In this example the associated Z value is 1.96. By symmetry of
the normal distribution the critical values are +1.96 and – 1.96
as shown in the figure below. From this we can now determine
the critical region (or rejection region) for our test.
CR > 1.96 and CR < –1.96
History
William Sealy Gosset is famous as a statistician who worked
for Guinness in the early 1900s, best known by his pen
name Student and for his work on Student’s t-distribution
which deals with the problem of small samples (n < 30).
Reject H0
Reject H0
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Overview
Students will be involved in conducting a hypothesis test
and then be required to relate the answers back to the
contextual problem.
Hints
Hypothesis testing of a population mean from a large
sample (n > 30) highlights the importance of the normal
distribution while at the same time drawing our attention
to a potential multitude of opportunities for using real life
applications involving the normal distribution.
Fail to Reject H0
-1.96
The conclusion that we can draw from the result in Step 5
might be something like, “Based on this we do not have
enough evidence to reject the null hypothesis. Therefore there
is no difference between the mean weight of tubing supplied
by Supplier A and the required company standard, i.e. Supplier
A’s tubing meets the required standard”.
1.96
Accurate sketches of statistic distributions can be generated on
this website : http://www.socr.ucla.edu/SOCR.html
Other statistical resources are available from the
Census at School website http://www.censusatschool.ie/