Pra U STPM 2022 Penggal 3 - Maths (T)

Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing
The choice of a one-tailed or a two-tailed alternative hypothesis depends on the conclusion to be drawn
if H is rejected. The position of the critical region can only be finalised once H has been stated.
1
0
For example, a manufacturer of electrical kettle conducts a test on the incoming heating element from
a supplier. He needs some criterion for deciding whether the average number of damaged elements has
increased, because he concerns about incurring an unnecessary financial loss. He sets up the hypothesis
that there is no change in damage rate and tests this against the alternative hypothesis that new shipment
received has higher defective rate. Such an alternative hypothesis will result in a one-tailed test with the
critical region located at the right tail.

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Example 4

A consumer group suspects that a local store’s 100 g packages of dried mangoes actually weigh less than
100 g. The group takes a random sample of 50 such packages and finds that the mean mass for the
sample is 99 g.
(a) State the null and alternative hypotheses for the hypothesis test.
(b) Determine whether hypothesis test is two-tailed test or one-tailed test and the location of the critical
region.

Solution: (a) H : µ = 100 g,
0
H : µ , 100 g.
1
(b) The group concerns only the mass of the packages less than 100 g and
thus test against the alternative hypothesis that the mass is inferior. Such
an alternative hypothesis will result in a one-tailed test with the critical
region falling in the extreme left tail of the distribution.




Exercise 5.1


1. A defendant who has been indicted for committing a crime stands trial in a court. Based on the
evidence, the judge will make a decision whether the defendant is innocent or guilty. State the null
and alternative hypotheses for the above court case.
2. A new drug is developed by a pharmaceutical company. It is the responsibility of the government to
5 judge the safety and effectiveness of this drug before allowing it to be sold to the public. State the null
and alternative hypotheses for this case.
3. A multiple-choice test consisting of 20 questions, each with five possible answers of which one is correct,
is given to a student. The purpose of this test is to determine the student’s familiarity with the subject.
A score of 8 or more correct answers will convince the examiner that the student has knowledge of
the subject being tested and is not simply guessing. Determine the null and alternative hypotheses for
setting up the hypothesis test.
4. A manufacturer of sports equipment has developed a new synthetic fishing line claimed to have a mean
breaking strength of 8 kg. If a random sample of 30 lines is tested and found to have a mean breaking
strength of 7.9 kg. Determine the null and alternative hypotheses for setting up the hypothesis test.
5. The recommended adequate intake of calcium for adults is 1000 mg per day. If we want to carry out a
hypothesis test whether the average adult on diet gets a daily intake of less than 1000 mg of calcium.
State the null and alternative hypotheses.



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Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing
6. There are two wrong conclusions from which you could draw in a hypothesis test: Type I and Type
II errors. Identify the Greek letter to denote the probability of each type of error.
7. The null hypothesis H : p = 0.35 is tested against the alternative hypothesis H : p . 0.35, where p is
0
1
a population proportion.
(a) Suppose that the decision procedure leading to nonrejection of the null hypothesis when in fact
it is false. What type of error is committed?
(b) If the decision procedure leading to the rejection of the null hypothesis when in fact it is true.
What type of error is committed?
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8. Consider the null hypothesis, H : A new teaching technique and the conventional classroom procedure
0
are equally effective versus the alternative hypothesis, H : A new teaching technique is either inferior
1
or superior to the conventional procedure. Describe the decisions taken that would result in Type I
and Type II errors if H is tested.
0
9. For the following statements, what happens to the likelihood that we reject the null hypothesis?
(a) The closer the value of a sample mean is to the value stated by the null hypothesis.
(b) The further the value of a sample mean is from the value stated in the null hypothesis.
10. Suppose one reads news stating that children in his country watch, on average, 5 hours of television
per week. In order to test this claim, he conducts a study on a random group of 30 children and finds
that the children in the group spend, on average, 4.6 hours watching television per week.
(a) State the null and alternative hypotheses.
(b) If the decision “reject the null hypothesis” is adopted, what decision error could be committed?
(c) Assume the decision “fail to reject the null hypothesis” is implemented, what decision error could
be made?
(d) State whether the test is a one-tailed test or a two-tailed test.

11. The normal curve for testing a null hypothesis H : µ = 50 is shown below.
0





0.025 0.025
–
x
25  = 50 75
Determine the
(a) rejection region, 5
(b) nonrejection region,
(c) critical values,
(d) significance level.

12. The normal curve for testing a null hypothesis H : µ = 32 is shown below.
0





0.01
–
x
31  = 32




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Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing
Determine the
(a) rejection region,
(b) nonrejection region,
(c) critical value,
(d) significance level.

13. It is given that H : μ = 70 is tested against H : μ > 70. State whether the test is a one-tailed test or
0
1
a two-tailed test. Assuming that the sample mean is normally distributed, sketch a normal curve and
indicate the rejection and nonrejection regions with a critical value arbitrarily chosen.
14. A hypothesis test is stated as H : μ = 15 versus H : μ ≠ 15. State whether the test is a one-tailed test
1
0
or a two-tailed test. Assuming that the sample mean is normally distributed, sketch a normal curve
and indicate the rejection and nonrejection regions with an arbitrarily critical value.
15. The pass rate of driving test in a country is reported to be 0.65. To test whether this claim is true, a
researcher selects a random sample of 20 driving test candidates. If the number of candidates passing
the driving test in the sample is anywhere from 9 to 17, the researcher decides not rejecting the null
hypothesis that p = 0.65; otherwise, he concludes that p ≠ 0.65. Use the binomial distribution to
determine the significance level of the test.
16. A record in a country reveals that the population mean life span of its people last year is 68 years
with a standard deviation of 7.7 years. A random sample of 50 recorded deaths in that country
during this year shows an average life span of 70.1 years. A critical region for the test statistic is such
–
that x > 70.1, find the corresponding value of the test statistic.



= 1 – 5.2 Testing Population Mean
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Evidence concerning the value of a population mean is provided by the sample mean. In the case where the
population variance is known, and a small sample is taken from a normal population or a large sample is
taken from any population, normal distribution is used to test a hypothesis about a population mean. In the
case where the population variance is unknown, a large sample is taken from any population, the hypothesis
test about a population mean can be carried out using a normal distribution as an approximation.

Population mean, variance known

We recall that the sampling distribution of the (sample) mean is normal for samples (any size) drawn from a
5 normal population, and is approximately normal for large sample drawn from any population. The sampling
2
2
distribution has mean µ – = µ and variance σ – = s n 2 , where μ and σ are the mean and variance of the
X
X
population from which we pick random samples of size n.
2
Suppose the population has unknown mean μ and known variance σ . Consider the hypotheses
H : μ = μ ,
0
0
H : μ ≠ μ .
0
1
Under the critical value approach, the significance level a is predetermined. The value of a corresponds to
the total area of the critical region.
–
X – µ
Under the null hypothesis, μ = μ , and the test statistic Z = σ 0 has a standard normal distribution,
0
N(0, 1). 
n
Thus, for a given a, the critical values of the random variable Z are –z a and z a . We have the probability
— 2 — 2
244





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Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing

–
1
2
P –z a , X – µ 0 , z a = 1 – a
σ
—
—
2
2

n
This expression can be used to indicate a nonrejection region for the null hypothesis H . Hence,
0
if –z a , z , z a , we do not reject H . On the other hand, if the calculated value of the test statistic falls in
—
0
—
2
2
the critical region, that is, z , –z a or z . z a , H is rejected.
2 — — 0
2
For a fixed significance level a, the critical regions and critical values are as shown in Figure 5.4.
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 (z)
1 – 

– 
2 –
2
z
–z  – 0 z – 
2 2
Figure 5.4
The following two examples illustrate how hypothesis tests are performed for the case in which the population
varience is known.
Example 5

A survey made by the Human Resource Ministry states that the average monthly salary of an executive
is RM4100 with a standard deviation of RM680. However, a sample of 25 executives selected recently
gives an average monthly salary per month of RM3850. Assuming that the average monthly salary of an
executive is normaly distributed, test, at the 1% significance level, whether the ministry’s claim is too high.

Solution: Let μ be the mean monthly salary of an executive claimed by the Human
–
Resource Ministry and x be the corresponding sample mean. Given information:
–
μ = RM4100, σ = RM680, n = 25, x = RM3850.
We are going to test whether the ministry’s claim of monthly salary is too high.
The significance level a is 0.01.
We carry out a hypothesis test using the following five steps. 5

Step 1 : State the null hypothesis and the alternative hypothesis.
H : μ = RM4100,
0
H : μ , RM4100.
1
Step 2 : Specify the significance level.
a = 0.01.

Step 3 : Select an appropriate probability distribution and determine the critical
region.
The population standard deviation s is known, the sample size is small but the
–
population distribution is normal. Hence, the sampling distribution of X is normal
σ
with mean µ and standard deviation  n . We will use the normal distribution
to perform the test.



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Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing

This is a one-tailed test with a critical region at the left tail. To locate the z value,
we look for 0.01 area in the normal distribution table. From the table, the z value
is approximately –2.33.
The critical region: z , –2.33.

Step 4 : Calculate the value of the test statistic.
–
x – µ
z = σ = 3850 – 4100 = –1.838
680
n
 
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25
Step 5 : Make a decision.
To make a decision, we compare the value of the test statistic to the critical value.
This value of z = –1.838 is greater than the critical value of –2.33 and thus it
falls in the nonrejection region. We do not reject H that the average monthly
0
salary of an executive is RM4100.
Note: Example 5: Normal population and small sample, known population variance.



Example 6

The length of a particular type of iron nails produced by a manufacturer has standard deviation 6.8 mm.
The target length for an iron nail is 38 mm. A supervisor takes length measurement of a random sample
of 100 nails and obtains a sample mean length of 39.4 mm. Test whether the mean length is on target.
Use the 5% significance level.
–
Solution: Let μ be the mean length of an iron nail and x be the corresponding sample
–
mean. Given information: μ = 38 mm, s = 6.8 mm, n = 100, x = 39.4 mm.
We are going to test whether the mean length of nails meets the target length
of 38 mm. The significance level a is given as 0.05.
The following are five basic steps in testing the hypothesis.
Step 1 : State the null hypothesis and the alternative hypothesis.
H : μ = 38 mm,
0
H : μ ≠ 38 mm.
1
Step 2 : Specify the significance level.
5
a = 0.05.
Step 3 : Select an appropriate probability distribution and determine the critical
regions.
The population standard deviation s is known and the sample size is large.
–
Hence, the sampling distribution of x is approximately normal with mean µ and
standard deviation σ . We will use the normal distribution to perform the test.

n
This is a two-tailed test with two critical regions, one at each tail. Since the total
area of the critical regions is 0.05, the area of the critical region at each tail is
0.025. To locate the z values, we look for 0.025 and 0.975 areas in the normal
distribution table. From the table, the z values are –1.96 and 1.96.
The critical regions: z , –1.96 and z . 1.96.





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Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing

Step 4 : Calculate the value of the test statistic.
–
x – µ 39.4 – 38
z = σ = 6.8 = 2.058
n
 
100
Step 5 : Make a decision.
To make a decision, we compare the value of the test statistic to the critical
value. This value of z = 2.058 is greater than the critical value of 1.96 and thus
it falls in the critical region. We reject H and conclude that the mean length of
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0
the iron nails produced by the manufacturer does not meet the target length of
38 mm.
Note: Example 6: Large sample, known population variance.


Population mean, variance unknown
2
In practice, the population variance σ is usually not known. However as long as the sample size is large, the
–
normal approximation for the sample mean X remains valid even if s is replaced by its unbiased estimate
–
^ X – µ .
σ. We can then apply the general test procedure using the test statistic Z =
^
σ
n

n
2
Note : σ = n s where s is the sample variance given by s = 1 ∑ (x – x) .
2
2
^ 2
– 2
n – 1 n i = 1
Example 7
A sample of 60 watches of a particular brand is checked for accuracy at 10:00:00 hours. Let μ denote the
true mean watch reading when the actual time is 10:00:00 hours. The resulting sample mean and sample
standard deviation are 10:00:01.2 hours and 1.8 seconds respectively. Use a significance level of 1% to
decide whether the evidence from the sample suggests the watches are fast.
–
Solution: Let x be the mean watch reading for the sample. Given information: μ = 10:00:00
–
hours, s = 1.8 seconds, n = 60, x = 10:00:01.2 hours.
2
^
s is calculated as s = n s
^
n – 1
= 60 × 1.8 2 5
59
= 1.815 seconds
We are going to test whether the mean watch reading is fast. The significance
level a is 0.01.
We carry out a hypothesis test using the following five steps.
Step 1 : State the null hypothesis and the alternative hypothesis.
H : μ = 10:00:00 hours,
0
H : μ . 10:00:00 hours.
1
Step 2 : Specify the significance level.
a = 0.01.
Step 3 : Select an appropriate probability distribution and determine the critical
region.



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Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing

The population standard deviation is not known and the sample size is large.
We will use the normal distribution to perform the test.
This is a one-tailed test with a critical region at the right tail. To locate the z
value, we look for 0.01 area in the normal distribution table. From the table, the
z value is approximately 2.33.
The critical region: z . 2.33.
Step 4 : Calculate the value of the test statistic.
–
x – µ
z = = 1.2 – 0 = 5.122
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σ ^ 1.815
n
 
60
Step 5 : Make a decision.
To make a decision, we compare the value of the test statistic to the critical
value. This value of z = 5.122 is greater than the critical value of 2.33 and thus
it falls in the critical region. We reject H and conclude that the true average
0
watch reading is faster than the actual time of 10:00:00 hours.


Relationship between hypothesis tests and confidence intervals

The hypothesis testing is very closely related to the estimation by a confidence interval. For the case of a
2
normal population with unknown mean μ and known σ , both hypothesis testing and confidence interval
–
estimation are based on the random variable, z = X – µ . It turns out that the testing of H : μ = μ against
σ
0
0

n
H : μ ≠ μ at a significance level a is similar to calculating a 100(1 – a)% confidence interval for μ. If µ
0
1
0
is outside the 100(1 – a)% confidence interval, then H is rejected at a, and if μ is inside the 100(1 – a)%
0
0
confidence interval, then H is not rejected at a.
0
Exercise 5.2
1. Determine the critical region and the critical values for z at a = 0.02 in the hypothesis testing:
H : μ = 50 against H : μ , 50.
0 1
5 2. Find the critical values x given that H : μ = 850, H : μ ≠ 850, σ = 36, n = 80. Use a = 0.05.
–
0 1
–
3. Given that H : μ = 250, H : μ ≠ 250 with n = 80, x = 248, σ = 10.7. State whether the null hypothesis
1
0
H should or should not be rejected at a = 0.02.
0
4. For the hypothesis test H : μ = 16.8, H : µ . 16.8, what is your conclusion at a = 0.1 if n = 45,
– 0 1
x = 17.5, s = 2.5?
5. In a random sample of 20 observations from a normal distributed population with standard deviation
–
s = 27.6, the sample mean is x = 110. Use a = 0.01 to perform the following hypothesis test:
H : μ = 120, H : μ , 120.
0 1
Next, change the sample size to 200 and perform the test again. Explain why you make different
decisions as the sample size increases.




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Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing
6. An electrical company manufactures light bulbs, their life times measured in hours, approximately
normally distributed. The company claims that the light bulb has a mean life time of 1000 hours with
a standard deviation of 86 hours. A customer suspects that the mean life time may be lower. He tests
a random sample of 35 light bulbs and finds that the average life time is 960 hours. At the significance
level of 5%, does the data provide evidence to conclude that mean life time of a light bulb is 1000
hours claimed by the company?
7. A manufacturer claims that the mean fat content of his hot dog is 10%. Assume that percentage fat
content to be normally distributed with standard deviation of 3%. A consumer group, concerned about
the fat content of hot dog, submits a random sample of 30 hot dogs to a laboratory for analysis. The
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laboratory result shows that the mean fat content of the hot dog is 12%. Carry out an appropriate
hypothesis test, using significance level of 1%, in order to advise the consumer group as to the validity
of the manufacturer’s claim.
8. 9 babies are born in a hospital in a particular day. The weights, in kg, of the babies are recorded as
follows:
2.9 3.2 3.0 2.8 3.4 2.7 3.1 3.0 2.8
Assume that the sample is from a normally distributed population with standard deviation 0.3 kg.
Carry out a test, at the 5% significance level, on the hypothesis that the population mean weight is 3
kg.
9. A car manufacturer advertises a car that has 20 km/l fuel consumption. A random sample of 30 cars
gives a mean petrol mileage of 18.7 km/l with standard deviation 3.81 km/l. Assume that the sample
is from a normal population. Test the manufacturer’s claim at the 5% significance level.
10. The random variable X has a normal distribution with standard deviation σ = 21. A random sample of
–
size 16 gives the sample mean x = 89. Determine a 90% confidence interval for the population mean.
Then, test H : μ = 95 against H : μ ≠ 95 using a = 0.1. Explain how this confidence interval can be
1
0
used to test the hypothesis.
11. Production records show that a machine makes coins with a mean diameter of 24.5 mm. An inspector
selects a random sample of 100 coins and obtains a mean diameter of 25.3 mm with a standard deviation
of 2.57 mm. Determine whether the machine slipped out of normal operation at the 1% significance
level?

12. The national mean cholesterol level is approximately 220 units. 50 patients with high cholesterol levels
(over 265) participate in a drug study and are treated with a new drug. After treatment the sample
mean is 235 and the sample standard deviation is 41. One question of interest is whether people taking
this new drug still have a mean cholesterol level that exceeds the national average. What conclusion
would you get from this study by using a significance level of 2%? 5
13. A random sample of 40 steel bars is taken from one of the production lines. It is found that the sample
mean and sample standard deviation are 28 kg and 4.5 kg respectively. Investigate the claim that the
mean mass of a steel bar is 30 kg. Use a significance level of 5%.
14. It is claimed that a car owner drives, on average, more than 25 000 km per year. To test this claim,
a random sample of 60 car owners are asked to keep a record of the distance they travel. Would you
agree with this claim if the random sample shows an average of 26 500 km and a standard deviation
of 7590 km? Use a significance level of 5%.
15. A production supervisor measures the filled volume of a random sample of 80 cans of mango
juice labelled as containing 350 ml. The sample has mean volume 348.2 ml and standard deviation
5.9 ml. Let μ represent the mean volume for all cans of mango juice recently filled by this machine.
The supervisor test H : μ = 350 against H : μ ≠ 350 at a significance level of 1%.
1
0
(a) Find the critical values in ml.
(b) Explain whether the mean filled volume differs from 350 ml?

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Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing

5.3 Testing Population Proportion



Often we have to perform a hypothesis test about a population proportion. Evidence concerning the value
of a population proportion is provided by the sample proportion. We shall discuss hypothesis tests about
the population proportion for small sample, where direct evaluation of binomial probabilities is required.
We shall also discuss hypothesis tests about the population proportion for large samples, using the normal
approximation to the binomial distribution.
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Population proportion, small sample
Let a parameter p represent the unknown proportion of a population that possesses a certain characteristic.
If an independent observation is randomly obtained from the population it can then be taken as having a
probability p of showing that particular characteristic. If a random sample of n observations is taken from
the population, the number of observations exhibiting the character of interest can be realised as a random
variable X obtained from a binomial experiment, that is, X ~ B(n, p).
When the binomial parameter p, success probability in a binomial experiment, is to be tested using hypothesis
testing procedure, we will consider that this parameter equals some specified value. A hypothesis testing
problem would then become testing the null hypothesis H that p = p against the alternative hypothesis
0
0
which may be one of the usual one-sided or two-sided alternatives: p , p , p . p , or p ≠ p .
0
0
0
The appropriate random variable on which we base our decision criterion is the binomial random variable
X. Values of X that provide significant evidence indicating the success probabilities are far from p will lead
0
to the rejection of the null hypothesis. Consider the hypotheses
H : p = p ,
0
0
H : p , p ,
1
0
we use the binomial distribution with p = p and q = 1 – p to determine P(X < x). The value of x represents
0
0
the number of successes in our sample of size n. If P(X < x) , a, we reject H as the result is significant
0
at the significance level a. Likewise, for the hypotheses
H : p = p ,
0
0
H : p . p ,
1
0
we obtain P(X > x). If this probability is less than a, we reject H . Lastly, for the hypotheses
0
5 H : p = p ,
0
0
H : p ≠ p ,
1
0
we calculate P(X < x) if x , np , or P(X > x) if x . np . If the probability is less than a , we reject H .
2
0
0
0
The steps for testing a null hypothesis about a proportion versus various alternatives are:
1 State H : p = p and H : p , p , p . p , or p ≠ p 0
0
0
1
0
0
2 Specify the significance level
3 Determine the critical region
4 Calculate the appropriate binomial probability
5 Make a decision
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Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing

Example 8

An airline claims that, on average, 6% of its flights are delayed each day. On a given day, of 20 flights, 2
are delayed. Test the hypothesis that the proportion of delayed flights is 6% at the significance level of 0.05.
Solution: Step 1 : State the null hypothesis and alternative hypothesis.
H : p = 0.06,
0
H : p . 0.06.
1
Step 2 : Specify the significance level.
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a = 0.05.
Step 3 : Select an appropriate probability distribution and determine the critical
region.
We have np = 20 × 0.06 nq = 20 × 0.94
= 1.2 = 18.8
Since np , 5, the sample size is small. We will use binomial distribution to
evaluate directly the probability.
This is a one-tailed test with a critical region falling in right side. The sample
proportion can be considered an outcome of a binomial experiment with
p = 0.06 and n = 20. All x values such that P(X > x) , 0.05.
Step 4 : Calculate the appropriate binomial probability.

We have x = 2 and n = 20,
20 20
x
P(X > 2) = ∑ 1 2 0.06 (1 – 0.06) 20 – x
x = 3 x
2 20
x
= 1 – ∑ 1 2 0.06 (1 – 0.06) 20 – x
x = 0 x
= 1 – 0.885
= 0.115
Step 5 : Make a decision.
To make a decision, we compare the value of the binomial probability to the
significance level. This value of 0.115 is greater than the significance level of 0.05
and thus it falls in the nonrejection region. We do not reject H and conclude
0
that there is insufficient reason to question the airline’s claim. 5


Population proportion, large sample

^
When the sample size n is large, the sample proportion p = x is approximately normally distributed with
n
mean and standard deviation equal to µ ^ = p and s ^ = p(1 – p) respectively.
p p n
Hence, for a large sample, we use normal distribution to perform a hypothesis test about the population
proportion p. The sample size is large if np . 5 and nq . 5. Then the test statistic is given by

p – p
^
Z = ,
pq
n


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Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing

For a two-tailed test at the significance level a, the critical regions are given as z , –z a and z . z a , whereas,
—
—
the critical region for a one-tailed test is either z , –z or z . z . 2 2
α
α
Example 9

A botanist has produced a new variety of hybrid rice grain that has better ability to resist stem borer than
other varieties. He knows that 82% of the seeds from the parent plants germinate. He claims the hybrid
has the same germination rate. 300 seeds from the hybrid plant are tested and 233 germinated. Test the
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botanist’s claim at the 2% significance level.
Solution: Let p be the proportion of seeds from the new hybrid plant germinated and p
^
be the corresponding proportion for the sample.

^
Given information: p = 0.82, n = 300, and p = 233 .
300
We are going to test whether the claim by the botanist is valid. The significance
level a is given as 0.02.

The following are the five steps in testing the hypothesis.

Step 1 : Formulate the null hypothesis and the alternative hypothesis.
H : p = 0.82
0
H : p ≠ 0.82.
1
Step 2 : Specify the significance level.

a = 0.02.

Step 3 : Select an appropriate probability distribution and determine the critical
regions.

We have
np = 300 × 0.82
= 246
5
nq = 300 × 0.18
= 54

Since both np and nq are both greater than 5, the sample size is large. We will
use the normal distribution.

This is a two-tailed test with two critical regions, one in each tail. Since the
total area of the critical regions is 0.02, the area of the critical region in each
tail is 0.01. To locate the z values, we look for 0.01 and 0.99 areas in the normal
distribution table. From the table, the z values are –2.33 and 2.33.
The critical regions: z , –2.33 and z . 2.33






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Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing

Step 4 : Calculate the value of the test statistic.
^
p – p
z =
p(1 – p)
n

= 0.7767 – 0.82
0.82(1 – 0.82)
300
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= –1.952
Step 5 : Make a decision.
We compare the value of the test statistic to the critical value. This value of
z = –1.952 is greater than the critical value of –2.33 and thus it falls in the
nonrejection region. We do not reject H which states that the rate of germination
0
for the hybrid plant is 0.82.




Example 10

A manufacturing company has submitted a claim that 85% of components produced by a certain process
are non-defective. A new process is introduced to lower the proportion of defective components below
the current 15%. In a sample of 100 components produced with the new process, 7 are defective. Is this
evidence sufficient to conclude that the process has been improved? Use the 5% significance level.

Solution: Let p be the proportion of defective components produced by the existing process
and p be the corresponding proportion for the new improved process. Given
^
^
information: p = 0.15, n = 100, p = 7 .
100
We are going to test whether the evidence is sufficient to justify the improvement.
The significance level a is given as 0.05.
The following are the five steps in testing the hypothesis.
Step 1 : Formulate the null hypothesis and the alternative hypothesis.
H : p = 0.15,
0
H : p , 0.15.
1 5
Step 2 : Specify the significance level.
a = 0.05.

Step 3 : Select an appropriate probability distribution and determine the critical
region.

We have
np = 100 × 0.15
= 15
nq = 100 × 0.85
= 85
Since both np and nq are both greater than 5, the sample size is large. We will
use the normal distribution.




253






05 STPM Math(T) T3.indd 253 28/10/2021 10:24 AM

Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing

This is a one-tailed test with a critical region located at the left tail. The area
of the critical region is 0.05. To locate the z value, we look for 0.05 area in the
normal distribution table. From the table, the z value is –1.645.
The critical regions: z , –1.645.

Step 4 : Calculate the value of the test statistic.
^ p – p
z =
p(1 – p)
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n

= 0.07 – 0.15
0.15(1 – 0.15)
100
= –2.240


Step 5 : Make a decision.
We compare the value of the test statistic to the critical value. This value of
z = –2.240 is smaller than the critical value of –1.645 and thus it falls in the
critical region. We reject H and this evidence is sufficient to conclude that the
0
process has been improved at the 5% significance level.






Exercise 5.3

1. In a random sample of 20 independent observations obtained from a binomial distribution, a student
wants to test H : p = 0.3 versus H : p ≠ 0.3. The student decides to reject H if X < 3 or X > 11.
1
0
0
0
Find P(X < 3) and P(X > 11).
2. Suppose that X = 13 is a number of occurrence from a sample of 20 independent observations obtained
from a binomial population. Test the following hypotheses at the 10% significance level.
H : p = 0.8
0
H : p , 0.8
5 1
3. A hunter claims that he hits 70% of the wildfowl he shoots at. On one day he guns down 6 of the 12
wildfowls he aims at. Using the 5% significance level for a hypothesis test, what is your conclusion on
this claim?

4. If there is no gender bias in trainee selection, then the pool of potential trainees is 50% male and
50% female. In a sample of 10 trainees, it is found that there are only two women trainees. Is there
evidence of gender bias in trainee selection? Use the 10% significance level for the hypothesis test.

5. A magazine claims that 45% of its readers do not trust an advertisement on a certain health food
product. In a poll of 20 randomly sampled magazine readers conducted two years later, 11 state that
they do not trust the advertisement. At the 5% significance level, is there evidence to support the claim
that the percentage of the readers against the advertisement has increased?




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Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing
6. 15% of mothers in a country who gave birth last year were under 20 years of age. A sociologist claims
that births to mothers under 20 years of age is decreasing. He selects a random sample of 25 births
this year and finds that 2 of them are to mothers under 20 years of age. Use a significance level of 1%
to test the sociologist’s claim.

7. At a certain college, it is estimated that around 30% of the students drive cars to class. In a random
sample of 20 college students chosen, 7 are found to drive cars to class. Determine, at the 5% significance
level, whether the 30% of the students driving to college is a valid estimation.

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8. Suppose that there are 38 occurrence from a sample of 60 independent observations obtained from
a binomial distribution. Find the z-value of the test statistic for the hypotheses test of H : p = 0.5
0
against H : p ≠ 0.5.
1
9. Suppose that X = 8 is an observation obtained from a random sample of size n = 25. The sample is
drawn from a binomial distribution. Consider the following test of hypothesis:
H : p = 0.4
0
H : p , 0.4
1
Compute an exact probability, P(X < 8), and compare it with the corresponding probability using a
normal approximation.

10. A construction firm claims that the job of covering housing floors with tiles is 80% completed in a
new housing estate. Would you agree with this claim if a random survey of new houses in this estate
shows that 32 out of 50 have floor job completed? Use a significance level of 5%.

11. A sample survey indicates that, out of 650 births, 327 are boys and the rest are girls. Do these figures
confirm the hypothesis that the sex ratio is 50 to 50? Use a significance level of 10%.

12. A club claims that it receives 12% responses from its mailing to club members. A random sample of
200 shows that only 18 members respond. Test the claim at a significance level of 10%.

13. A study estimates that 47 per cent of people in a city wear spectacles. However, you believe the
proportion of people in the city who wear spectacles is less than 47 per cent. You conduct a study
and find that, among 250 randomly selected people in the city, 114 of them wear spectacles. Test, at
the 5% significance level, the estimation that 47 per cent of people in the city wear spectacles.

14. A patented medicine claims that it is effective in curing 80% of the patients suffering from cold. From
a sample of 100 patients using this medicine, it is found that only 72 are cured. Determine whether 5
the claim is valid at the 5% significance level.

15. In a controlled laboratory experiment, scientists discovers that 35% of a certain rats subjected to a
cancer-causing chemical injection later develop cancerous tumours. Would we have reason to believe
that the proportion of rats developing tumours when subjected to this injection has increased if a
repeated experiment finds that 26 of 60 rats develop tumours? Use the 5% level of significance.

16. A random sample of 120 recent donations at a certain blood bank reveals that 55 are of type O
blood. Does this suggest that the actual percentage of type O donations differs from 38%, the
percentage of the population having type O blood? Carry out an appropriate hypothesis test using
a significance level of 10%. Would your conclusions have been different if a significance level of 5%
had been used?




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Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing

Summary



1. A hypothesis test or significance test is a method of using sample data as evidence to test a statistical
hypothesis about a population parameter.

2. A null hypothesis is a statement about a population parameter that is assumed to be true until it is
rejected with strong evidence obtained from a sample. An alternative hypothesis is a statement about
a population parameter that will be true if the null hypothesis is rejected.
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3. A test statistic is a random variable whose value is used to determine whether a null hypothesis is
rejected in a hypothesis test.
4. A critical region is the set of values that leads to the rejection of the null hypothesis in favour of the
alternative hypothesis.

5. (a) Type I error occurs if the null hypothesis is rejected when the null hypothesis is true. The
probability of making this error is called the significance level of a test denoted by a.
(b) Type II error occurs if the null hypothesis is not rejected while in fact the null hypothesis is
false. The probability of making this error is denoted by b.

6. A hypothesis test which has one sided critical region in either left or right tail is called a one-tailed
test. A hypothesis test which has two critical regions, each at the left tail and right tail, is called a
two-tailed test.

7. The general test procedure is as follows:
• State the null and alternative hypotheses
• Specify the significance level
• Select an appropriate probability distribution and determine the critical region(s)
• Calculate the value of the test statistic
• Make a decision
–
8. To test a hypothesis about a population mean with known variance the test statistic is Z = X – µ ,
σ
where the population is normal if the sample size is small. 
n
9. To test a hypothesis about a population mean with unknown variance where the sample is large, the
–
5 test statistic is Z = X – µ .
^
σ

n
10. To test a hypothesis about a population proportion, where the sample size is small, the critical region(s)
is (are) as follows:
(a) All x values such that P(X < x) , a for p , p 0
(b) All x values such that P(X > x) , a for p . p 0
(c) All x values such that P(X < x) , a when x , np and all x values such that P(X > x) < a
2
2
0
when x . np for p ≠ p
0 0
11. To test hypothesis about a population proportion, where the sample size is large, the test statistic is
p – p
^
Z = .
p(1 – p)
n
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Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing

STPM PRACTICE 5


1. A random sample of 70 observations taken from a normal distributed population, with standard
–
deviation s = 7.2, gives the sample mean x = 60.8. Test, at the 5% significance level, H : μ = 60 against
0
H : μ . 60.
1
2. A random sample of 103 observations is taken from a certain population. It is found that its sample
mean is 189 with a standard deviation of 29.7. The null hypothesis H : μ = 200 is to be tested against
0
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H : μ ≠ 200. If the test is performed at the 2% significance level, what conclusion do you draw?
1
3. Suppose a random sample of 20 independent observations is obtained from a binomial distributed
population and the number of successes is 8. Use a significance level of 1% to test H : p = 0.20 against
0
H : p ≠ 0.20.
1
4. It is given that the number of successes, x = 85 and the number of independent observations,
n = 300 for a random sample obtained from a binomial distributed population. Use a significance
level, a = 0.05 to perform the hypothesis test H : p = 0.32 versus H : μ , 0.32.
1
0
5. A person claims that the weather forecasts by a meteorologist are no better than the outcomes of tossing
a fair coin. If a head is obtained then there will be no rain, and if a tail is obtained then there will be
rain. He records the weather for 50 randomly chosen days. The meteorologist forecast is correct on
34 of these days.
(a) Write the hypotheses clearly.
(b) Use a significance level of 1% to test the claim of the person.

6. An environmental department wants to determine whether a cleanup project at a lake has been effective.
This is to be done by recording dissolved oxygen content (in ppm, parts per million) in the lake, with
higher values indicating less pollution. Prior to the cleanup project the mean dissolved oxygen readings
around the lake is reported as 9.80. Six months after the initiation of the cleanup, a random sample
of 80 readings gives the mean and standard deviation as 9.95 ppm and 0.51 ppm respectively.
(a) State null and alternative hypotheses for a hypothesis test.
(b) Carry out the test at a significance level of 5%.
7. A shopkeeper realises that 20% of customers buy a drink from the storage. During the renovation of
the shop a new storage was installed. He picks a random sample of 20 customers and finds that only
1 customer have bought drinks from the new storage. Using a significance level of 5% to test whether
there has been a change in the proportion of customers buying a drink from the storage.
5
8. A real estate agent claims that 60% of all apartments being built today are 3-bedroom units. To test
this claim, a sample of 50 new apartments is inspected and it is found that the proportion of these
apartments with 3-bedroom units is 75%. Perform a hypothesis test at a significance level of 2%.
9. The contents of a random sample of 9 containers of a particular paint are 5.2, 4.7, 4.6, 5.3, 5.1, 4.8,
4.9, 5.4, and 4.8 litres. Assume that the contents is normally distributed with standard deviation 0.2
litre. Use a 10% significance level to determine whether the mean content of the containers is 5 litres.

10. The mean height of male students in a certain college has been 164.9 centimetres with a standard
deviation of 14.3 centimetres. Is there strong reason to believe that there has been a change in the mean
height of male students if a random sample of 90 male students in the college has a mean height of
168.5 centimetres? Assume that the sample standard deviation remains the same and use a significance
level of 5%.




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Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing
11. An exit poll by a news station of 800 people in a certain parliamentary constituency finds 390 voting for
candidate A. Does the data support the hypothesis that candidate A receives 50% of the parliamentary
constituency’s votes at a significance level of 10%.

12. According to an article published 2 years ago, the mean number of cigarettes smoked per day by adults
who were daily smokers was 11.6. To determine whether adults who are daily smokers nowadays smoke
less than the general population of daily smokers in the past, a random sample of 100 adults who are
current daily smokers and record the number of cigarettes smoked on a randomly selected day. The
data give a sample mean of 10.8 cigarettes and a standard deviation of 3.9 cigarettes. Perform, at the
5% significance level, a test to determine whether current adults who are daily smokers smoke less
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than the general population of daily smokers two years ago.
13. A pizza shop in a district advertises that their average delivery time to the district is at most 15 minutes.
A sample of 60 delivery times has an average of 18 minutes. The true standard deviation of delivery
times is 8 minutes. Is there sufficient evidence to reject the shop’s claim at the 5% significance level?
14. For 150 bottles labelled as “350-millilitre”, the average amount of soft drink filled by the machine is
348.5 millilitres. The population standard deviation is known to be 11.5 millilitres. If the expected
amount (per bottle) of the soft drink filled by the machine is not equal to 350 millilitres, then the
machine is “out of control”, and must be shut down for repairs. Test whether the machine is out of
control at the 5% significance level.

15. A certain type of medicine is known to be 60% effective in relieving an arthritis pain. To determine
whether a new and somewhat more expensive patent medicine is superior in alleviating the arthritis
pain, a sample of 500 patients is taken and it is found that the new medicine provides relief of 318
patients. Carry out a test at a significance level of 5%.

16. The hourly french fried potato output by a certain brand fry machine is advertised to be 60 kilograms.
For the new machine purchased by a fast food restaurant, tests are run for 50 different one-hour periods,
producing an average output of 57.2 kilograms, with a standard deviation of 6.8 kilograms. At the 5%
level of significance, does the fast food restaurant management have grounds for complaints?
17. A random sample of 80 observations obtained from a population produces the results:
80 80
2
∑ x i = 2880, ∑ x i = 129 276.
i = 1 i = 1
(a) Calculate an unbiased estimate of the population mean and variance.
(b) Determine a 95% confidence interval for the population mean.
(c) Test, at the 5% significance level, H : μ = 34 against H : μ ≠ 34.
0
1
5 (d) Explain the relationship between the confidence interval obtained in (b) and the result of the test
in (c).
18. The mean lifetime of a sample of 200 picture tubes produced by a manufacturer is found to be
14 800 hours with a standard deviation of 1060 hours. Use a significance level of 5% to test, on the
basis of this data, whether the mean lifetime of all such tubes made is 15 000 hours as claimed by
the manufacturer.
19. In a sequence of rolling a dice 500 times, ‘ones’ are obtained 77 times. Test, at a significance level of
10%, the hypothesis that the dice is unbiased.
20. A traffic authority is concerned with the increasing traffic congestion in a city. The authority knows
that the mean time per journey between the city centre and the nearby town during the peak hours
was 32.5 minutes with a standard deviation of 19.7 minutes five years ago. This year, for a random
sample of 150 car drivers, the mean travelling time is 36.8 minutes. Does the authority justify the
worry about the traffic congestion getting worst this year if a significance level is chosen to be 5%?



258






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Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing
21. A large chain of telecommunication service provider introduces a training programme for counter staff
to increase their productivity and morale. The management believes that the mean time for collecting
customers’ phone-bill payment should be 120 seconds. After the training programme, the times to
collect 90 bills are found to have a mean of 113.5 seconds and a standard deviation of 35 seconds.
(a) Determine, at the 5% significance level, whether, after the training programme, the mean time
to collect a bill is less than 120 seconds.
(b) What assumption is necessary for the test in part (a) to be justifiable?

22. The proportion of members of a certain badminton club who are able to explain the rules of badminton
game correctly is p. A random sample of 10 members of the badminton club is selected and 4 members
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are able to explain the rules correctly. Test the null hypothesis, H : p = 0.7 against the alternative
0
hypothesis H : p , 0.7 at the 10% significance level.
1
23. For a binomial distribution B(15, p), it is to test H : p = 0.25 versus H : p . 0.25.
0
1
(a) Using a significance level of 5%, determine the critical region for this test. The area of the critical
region at the left end should be as close as possible to 0.05.
(b) Find the actual significance level of this test and the critical value.
(c) Draw a conclusion from this test.
24. In a class, 25 students are selected randomly and their reaction times, in seconds, to a particular
experiment, are measured. The mean reaction time for the students is 6.2 seconds with a variance of
2
4 s .
(a) Find the 95% confidence interval for the mean reaction time for all students.
(b) State any assumptions necessary to make this valid estimate.
(c) Based on your answer in (a), would the null hypothesis that the true mean is 7.0 seconds be
rejected at the 5% significance level? Why?
25. A peanut factory claims that at most 6% of the peanut shells contain no nuts. 100 peanuts are selected
at random and it is found that 9 of them were empty. Test at the 5% level of significance whether or
not the claim made by the factory is true.
26. Scores from a standardised memory test of all the secondary students in a school are normally distributed
with a mean μ and a standard deviation σ = 18. The score of a random sample of 60 such students has
a mean of 76. Perform a test to determine whether μ is less than 80, using the 5% significance level.
27. A random sample of 65 bags of groundnut which are labelled as 100 grams, are weighed. The mean
weight of the bags is 99 grams with an estimate of the standard deviation of 4.21 grams. Test whether
the mean weight of all bags of groundnut packed is less than 100 grams. Use the 5% level of significance.

28. A machine fills bottles with cooking oil. Prior to maintenance of the machine, the volume of cooking
oil in a bottle could be modelled by a normal distribution with mean 554 ml and standard deviation 5
3.5 ml. Following this new setting of the machine, the mean volume of cooking oil of a random sample
of 12 bottles is 555.1 ml.
(a) Carry out a test, at the 10% significance level, to decide whether the mean volume of cooking oil
filled by the machine has changed. Assume that the distribution of volume is still normal with
the variance remain unchanged.
(b) Find the largest significance level such that there is evidence that the mean volume of the cooking
oil has increased.
29. Steel rods produced on a production line are supposed to average 25.2 mm in diameter. Assume the
diameters of the steel rods are normally distributed with standard deviation 0.11 mm. It is desired to
check that the mean diameter of the rods is under control within certain limits. Suppose a random
sample of 80 steel rods is taken and the diameter of each rod is measured. Test at the 2% level to
determine the limits of the mean diameter such that the steel rods produced would be acceptable.




259






05 STPM Math(T) T3.indd 259 28/10/2021 10:24 AM

ANSWERS





1 Data Description 3. (a) 6.3 hours, [6.0 – 6.4] hours
(b) 0.63 g, [0.60 – 0.64] g
Exercise 1.1
1. (a) discrete (b) continuous 4. (a) Girls Boys
(c) continuous (d) discrete 0 9
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XX (e) discrete 0 11
1 13
2. (a) Stem Leaf 1 1 1 1 0 0 15 1 1
4 4 1 1 17 0 0 1 1
5 1 0 0 19 0 0 0
6 0 0 6 6 1 1 0 21 0 1 1 1 1 1 1 1
7 2 3 5 9 9 0 0 0 23 1
8 1 1 3 4 5 6 6 7 7 25 0 0
9 1 3 Key: 1|15 Key: 21|1
10 5 means 16 means 22
Key: 7|2 means 7.2 mm Comment: Boys hold their breath longer than girls.

(b) Stem Leaf (b) Chemistry Mathematics
45 2 4 1 3 2
50 2 2 4 6 6 5 4 3 4 2 4 5
55 3 4 4 9 8 7 3 3 0 5 2 7 9
60 0 2 2 3 3 8 7 2 1 6 2 4 5 6 7
65 0 0 1 2 3 4 4 7 4 7 1 3 3 7 8 9
70 0 1 2 4 4 2 8 4 6
75 1
80 1 Key: 6|4 Key: 5|2
means 46 means 52
Key: 50|2 means 52 kg
Comment: Students achieved higher marks in
(c) Stem Leaf Mathematics compared to Chemistry.
1 2 2 2 2 Exercise 1.2
1 5 5 5 5 5 5 1. (a) f
1 6 24
1 8 8 8 8 9 9 9 9
2 0 0 1 1 1 1
2 2 2 2 20
2 4 4
16
Key: 1|9 means 0.19 second
12
(d) Stem Leaf
0 2 3 8
0 8
1 2 4
1
2 1 4 0 x
8 8 9 5 6 7 8 9 10
3 0 1 2 (b) f
3 5 6 7 8 8 9 40
4 2 2
4 8 9 30
5 2
5 6 6 8 20
6 2
6 5 10
Key: 4|8 means 4.8 0 x
4.5 9.5 14.5 19.5 24.5 29.5 34.5 39.5

285






Ans STPM Maths(T) T3.indd 285 28/10/2021 10:25 AM

Mathematics Semester 3 STPM Answers

2. (a)
250 20
Number of applicants 150 Number of workers 24 8 4 0
16
200
12

100
Net income (RM)
400 600 800 1 000 1 200 1 4001 600 1 800
50
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Exercise 1.3
1. (a)
0
5 10 15 20 25 30 35 40
Age (years) 50
(b) 200 166 40
Number of workers 150 48 92 65 31 1 2 Cumulative frequency 20
30
100
50
10
0
2468 12 12 20 3.2 40 0 x
2 4 6 8 10 12 14
Distance (km)
(b)
3. 16 15 50
Number of patients 12 8 4 12 9 6 4 5 1.5 Cumulative frequency 40

30
0
7.0 10.5
21.0 24.5
14.0 17.5
28.0 31.5
3.5
Time (minutes) 35.0 38.5 42.0 45.5 49.0 20
10
4. (a) 20 0 0.5 10.5 20.5 30.5 40.5 50.5 60.5
Number of workers 24 8 2. 120 Time (minutes)
16
12
0 4 Number of days 80
40
400 600 800 1 000 1 200 1 4001 600 1 800
Net income (RM) 0
4.5 9.5 14.5 19.5 24.5 29.5 34.5 39.5
Number of crime cases
(b) Net income (RM) Number of workers
‘Less than’
400  x  600 5 Upper boundary cumulative frequency
600  x  800 11  0.5 0
800  x  1 000 23  4.5 5
1 000  x  1 200 6  9.5 42
 14.5 129
1 200  x  1 400 18
 19.5 250
1 400  x  1 600 9
 24.5 327
1 600  x  1 800 11
 29.5 369
The two histograms drawn in (a) and (b) are the  34.5 390
same except the income classes in (b) are RM200 less  39.5 400
than the one in (a).




286






Ans STPM Maths(T) T3.indd 286 28/10/2021 10:25 AM

Format: 190mm X 260mm Extent= 312 pgs (15.54 mm) 70gsm Status: Cover Master BM vervion_2nd imp

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