Title : “How does different sources of river water affect the algal bloom of Pediastrum boryanum a case study for three months in Segamat,Johor?”

1. 1
1.0 INTRODUCTION
Water is claimed as polluted when there are presence of substance that can give bad effect
on the water and the organism that depend on the water (Simons & R. Wellnitz,
2002).One of the most serious human pollution impacts has been the enrichment of
freshwater with phosphate, which promotes the growth of harmful populations of Elodea
(waterweeds), Pediastrum boryanum (green algae), and cyanobacteria (blue-green algae).
This process, which converts clean lakes into polluted ones, is known as eutrophication
(E.Graham, M.Graham, & W.Wilcox, 2003). Oligotrophic is a condition where the clean
and clear water are found (can be stream or lakes). These waters are defined by low
mineral nutrient content and low primary productivity (the amount of carbon fixed in
photosynthesis per unit area per year), but high species diversity. Although many
different kinds of phytoplankton and other organisms may occur in oligotrophic waters,
they are few and far between. Such waters produce desirable game fish and do not
generate harmful algal blooms or excessive growth of weeds (E.Graham, M.Graham, &
W.Wilcox, 2003). Meanwhile, eutrophic waters have a high mineral nutrient content and
primary productivity, but low species diversity. Eutrophic waters are not known for good
fishing unless they have been artificially stocked with fish, and people regard them as
degraded waters having low quality. Eutrophic freshwaters may be crowded with fast-
growing, weedy plants that may entangle swimmers, interfere with boat traffic, and shade
out phytoplankton in the underlying waters.
There are many types of pollutant such as poisonous chemicals and metals, radioactive
substance, toxic, pesticides and fertilizer such as nitrates and phosphates. When these
pollutants dissolve in water and absorbed by algae, these can lead to dangerous food
chain. This is because, when the contaminated algae are eaten by small water animal, the
small water animal will be eaten by frogs or fish. When the frogs or fish eat the small
water animal, they also consume the chemical contain in the algae. When human eat the
fish, the dangerous chemical build up in their body too. This may lead to the birth defects
or illness such as cancer as the long-time pass. Moreover, the pollutants can affect human
in many ways such as drinking impure water, eating contaminated fish, bathing and
swimming in polluted water as the polluted water can irritate the skin and cause serious
problems such as skin cancer.

2. 2
Nowadays, farmers usually spray the pesticides and fertilizer on their field to produce
better crops. When rains fall, the rain will carry with them the chemicals on the field as a
runoff. This runoff will enter the pond, lake of rivers and creates a mixture of water-
chemical. Naturally, the pond changes over time as the eutrophication take place. The
more plants grow in the water, the more the dead plant piles up on the bottom, hence the
pond become shallower and warmer. As the plant matter decays increases, the oxygen
levels in the water decreases. In addition, when the fertilizer contents in the pond
increases, this process will speeds up. The thick layers of scum form on the surface of
pond water and this prevent the sunlight to penetrate the base of the pond, hence retarded
the process of photosynthesis of the base pond plant. The died plant is decomposed by
bacteria that undergoes aerobic respiration. As the time past, more plant will die because
lack of oxygen.
Even though phytoplankton algae are small but they are big enough to be seen though a
microscope (E.Graham, M.Graham, & W.Wilcox, 2003). They able to grow on the
surface of water by manipulating the water motions that suspend and circulate them. This
process exposes phytoplankton to a constant supply of dissolve mineral nutrients, which
they absorb directly from the water. Some features that contribute the survive of the
phytoplankton are the present of flagella, spines, formation of large colonies of cells, and
coatings of mucilage. These features help them swim or stay afloat and defend
phytoplankton from some type of hungry aquatic animals.
One of the characteristics of eutrophic water is they are cloudy because it is concentrated
of suspended cyanobacteria. Lakes having persistent and enduring cyanobacterial blooms
are described as hypereutrophic. When cyanobacteria become very abundant, about 70%
of the time they produce toxins that are released into the water and can kill waterfowl,
together with wild and domestic animals. Domestic animals should not be allowed to
drink water having a visible blue-green bloom, and people should limit their exposure
when swimming or water-skiing in such waters.

3. 3
1.1 RESEARCH QUESTION
My research question “how does different sources of river water affect the algal bloom of
Pediastrum boryanum, a case study for three months in Segamat,Johor?” . There are two
different sources of river water that will be examine which are one with the source from
the vegetable farm and another one is source from the domestic sewage. In order to
determine which river is more polluted, I will look at the oxygen level contain in the
water samples from both rivers. This is because, the cleanliness level of water usually is
depends on their enrichment of oxygen which is a very important element for an organism
to sustain their life especially for the aquatic life. The technique that I will use is by
measuring the oxygen level content in the water in five different places. This requires the
0.1% methylene blue solution as indicators for the oxygen level content, which is the
faster the blue colour solution decolourise, the less the oxygen level contain in that
particular water samples. The time taken for the 0.1% methylene blue solution to
decolourise is measure by using digital stopwatch. The significance of this study is to
know which factor affects the most massive growth of Pediastrum Boryanum so that
people can aware the factor and reduce the usage of that particular factor. Hence, the
pollution can be reduced.

4. 4
2.0 VARIABLES
The independent variable for my research is the sources of the river water. In this experiment,
comparison of two rivers that each river have different factor to affect algal bloom. In river
A, the source of water is originates from the domestic sewage.
Figure 1 show river A (Sungai Kenawar) which has no algal bloom on its surface. The source
of this river originates from the domestic sewage.
Meanwhile, in river B, the source of water is originates from the vegetable farms. This farm
consist of various type of vegetables such as brinjal ( Solanum melogena) , spinach (Spinacia
oleracea) and sweet nail . These types of vegetables consume commercial fertilizers which
consist of fertilizers nutrient ratio which is commonly used by the gardener is a fertilizer with
a ratio of 5-10-5 (Nardozzi). This ratio indicates of the ratio of nitrogen, phosphate and
potash and it is considered as a complete fertilizer because it contains all nutrients needed by
the plants to grow. On the surface of the water of river B, it can be seen a layer of green
algae.

5. 5
Figure 2 show the present of algal on the surface of the river B (Sungai Kapeh).The source of
this river is originates from the vegetables farm that consumes commercials fertilizers.
Figure 3 show the three types of vegetables that are planted in the vegetable farms. From the
left, image A is brinjal (Solanum melogena), image B is spinach (Spinacia oleracea) and
image C is swamp cabbage (Ipomoea aquatica). It can be seen that on the soil, there are
fertilizers on it.
Five samples in three trials in three different points along the river are taken. This water
samples is then kept in the reagent bottles to make sure there is no additional of oxygen or
oxidation process occur.
A B C

6. 6
The dependent variable for this research is the growth of Pediastrum Boryanum. Massive
growth of Pediastrum Boryanum can be influenced by certain factor such as nourish of
nutrition from certain fertilizers (commercial fertilizers). The collected samples of river water
are then tested by 0.1% methylene blue solution as the indicator of the cleanliness of the
water. This method is known as Biochemical Oxygen Demand (BOD). The higher the BOD,
the lower the concentration of dissolve oxygen in that water. Hence, the water is claimed as
polluted. The time taken for the 0.1% methylene blue solution takes to decolourise is noted as
the methylene blue solution indicates the presence of oxygen molecule. The faster the time of
the 0.1% methylene blue solution takes to colourless, the lower the concentration of dissolve
oxygen in that solution.
The variables that are being controlled are the concentration of the methylene blue solution
and the volume of water samples. The concentration of methylene blue solution used is 0.1%
in all experiment. The volume of water samples also controlled by making sure that the 50 ml
of water samples are used in each reagent bottles. These two variables should keep controlled
as it may effects the result of the experiment if it were not controlled.
The uncontrolled variables in this experiment are the environment of the cupboard that stores
the reagent bottles (containing water samples). Even though the cupboard is covered by using
blinds, there are some light that could still penetrate the cupboard and hits the reagent bottles.
There are several safety precautions that have to take serious while doing this case study.
First, take the water samples possibly at the centre of the river. To do this, I have to make
sure the real depth of the river by using a long wooden stick. I insert the wooden stick inside
the centre of the river and take the reading of the highest water level. thus, I can be aware
about the depth of the river while taking the water samples and make preparation such as hold
a rope that been tightly tied to the adjacent trees. Next, I have to wear a pair of thick soled
shoe because there are chances of the base of the river contains sharp object such as off
stump, rock and broken glass. By wearing this shoe, I will keep my leg safe from any
accidental injuries. Moreover, the river water is cloudy and has limited sight to the bottom of
the river.

7. 7
3.0 HYPOTHESIS
The alternative hypothesis is the time taken for the 100 ± 0.1ml of 0.1 % methylene blue
solution to decolourise in river B is shorter than river A. This is because in river B, it contains
the nutrient supplied from the dissolved fertilizers from the vegetable farm that nourish the
growth of Pediastrum Boryanum. This algal will cover the surface of the river. Hence prevent
the penetration of sunlight to the bottom of the river. So, the aquatic plants unable to undergo
photosynthesis and no oxygen are produced. As the time passes, the aquatic plant died, then
the decomposers will decompose the dead plant and this process required oxygen. Thus, the
dissolve oxygen concentration decreases and finally the river are no longer suitable for other
aquatic life to live there.
The null hypothesis for this experiment is time taken for the 100 ± 0.1ml of 0.1 % methylene
blue solution takes to decolourise is the same for river A and river B.
4.0 LIST OF APPARATUS AND MATERIALS
The apparatus used in this experiment are 30 of 250 ml reagent bottles with stoppers. This
reagent bottles need to be not leaking. This is because the reagent bottles will used to store
the water samples from the river water. The stopper used to be functioning well as it must
prevent the diffusion of atmospheric oxygen with the water samples.
Syringes are used when inserted the 100 ± 0.1ml of 0.1 % methylene blue solution in the
water samples. The syringe should be calibrated as it may establish the reliability of the
instrument. It is used in this experiment as the 100 ± 0.1ml of 0.1 % methylene blue solution
must be inserted in the water samples softly and to avoid any bubbles formed as it may
increase the dissolved oxygen level and may affect the results.
Filter funnel is used when I want to pour the water sample from the beaker into the reagent
bottles. As the mouth of the reagent bottles are quite small compared to the beaker, it is easier
to use filter funnel. Thus, this will avoid the spill of the water samples.
Next, six stopwatches with absolute uncertainty of ±0.1 seconds are used to take the time
taken for the 100 ± 0.1ml of 0.1 % methylene blue solution decolourise. Then, a marker pen
and a labelling sticker are used to labelled the reagent bottles to avoid confuses. A 500 meter

8. 8
measuring tape with absolute uncertainty of ±0.1 cm is used to measure the distance between
three points along the two rivers.
About 100 ± 0.1ml of 0.1 % methylene blue solution is used for each reagent bottle in order
to test the oxygen level in the river water. this methylene blue solution is used because it is
the most suitable indicator as it able to indicate the presence of oxygen which is it is blue in
colour if there are the presence of oxygen, meanwhile it decolourise when there is no
presence of oxygen.
50 ml of water samples is used to be test. This water samples is filed in the reagent bottles
and then tested with 100 ± 0.1ml of 0.1 % methylene blue solution and the time taken for the
methylene blue solution to decolourise is taken.

9. 9
5.0 METHODOLOGY
About 45 field trials were carried out at Kenawar River, Segamat, Johor (2°30'12.7"N and
102°50'51.2"E) between May and July in 2015. The site is a typical river where the
domestic sewage and fertiliser are drawn into it. Manual sampling method is used in order
to get the water samples in three different points along the river which is 50 m distance
between each point.
Figure 4 shows the measuring distance of 50 m between each point of the water samples
were taken. This procedure is conducted the same way for both rivers.
Each point is named as point A, B and C respectively. Manual grab sampling technique is
used as it is the standard method for collection of base flow samples (Migliaccio, Harmel,
C. Smiley, & Jr, 2011). In preparation step, 15 reagent bottles were used in order to
collect the water samples. The reagent bottles are cleaned up first to prevent any
microorganisms intercept during the experiment. A dropper is test to make sure it is
functioning well and no leakage occurs. This is to prevent any leakage of methylene blue
solution when added to the water samples.100 ± 0.1ml of 0.1 % methylene blue solution
is measured and placed in a dark reagent bottle with stopper.

10. 10
Figure 5 shows the 15 reagent bottles containing water samples for a point in each river.
The river water was flow fluently and there are algae on the surface of the water. A start
point is marked by using a cone. By using a meter tape, measure 50 meter distance
between the start point to the left and mark using a cone. Then, measure 50 meter distance
between the start point to the right and mark using a cone. This can simply be understood
from figure 6 below.
Figure 6
The advantages of using grab sampling are it is relatively safe, simple, and inexpensive,
can be performed at any location, and is not subject to equipment theft or vandalism but it
provide limited information about the condition of water samples (Migliaccio, Harmel, C.
Smiley, & Jr, 2011).However, manual grab sampling is the best choice to reduce the cost
of purchasing automatic sampling equipment. At point A, collect the water sample by
using a bail.

11. 11
Figure 7 shows the preparation of the bail in order to undergo the manual grab sampling
technique. The bail is used to collect the water sample from the river.
Place 50±0.1 ml of water from the same source (river water) into each of 5 reagent
bottles. Close the reagent bottles tightly by using stopper. Place the 5 reagent bottles in
dark cupboard. This is to prevent any autotroph to undergo photosynthesis. So that the
dissolved oxygen level is not increased from the original level after the taking samples
process. The same process goes to the point B and Point C.
To test the quality of the water sample, an indicator solution which is 0.1% methylene
blue solution is used in each 45 trials.
Figure 8 shows the addition of 100 ±0.1 ml of 0.1% methylene blue solution into the
water samples.

12. 12
Methylene blue solution is used to test the presence of oxygen in the water samples.
When it is blue, it indicates there is oxygen and when the solution decolourise, it indicates
that there is no oxygen. Measure 100 ±0.1 ml of 0.1% methylene blue solution by using
measuring cylinder and suck it into the syringe. The methylene blue solution is added at
the base of the water samples. To make sure of that, syringe with needle is used in
addition of methylene blue solution into the water samples. Make sure that the needle of
the syringe is at the base of the water sample but do not until it touch the bottom of the
reagent bottle. This is to avoid the methylene blue solution being oxidised by atmospheric
oxygen. The reagents bottles are closed as quickly as possible by using stopper and don’t
shake the reagent bottles. This is to avoid aeration or to prevent more oxygen from
dissolving in the water samples (Yeat, 2013). Start the stopwatch simultaneously and
place all the 15 reagent bottles in a dark cupboard. The colour changes of the water
samples are observed from time to time. The time taken for the 0.1% methylene blue
solution to become colourless is observed and recorded in the table.

13. 13
6.0 DATA COLLECTION
Refer the appendix 1 and 2 for the quantitative data of river A and river B.
Refer the appendix 3 for the qualitative data of river A and river B.
River A River B
Total number of samples, n 45 45
Mean, 𝑥̅ 88.1 65.7
Standard deviation, S 0.7 8.1
Table 4 shows the mean, standard deviation and total number of samples in river A and B.
Degree of freedom 88
t value 18.6
Critical value 1.99
Table 5 shows the t test data.

14. 14
7.0 DATA ANALYSING
7.1 Calculating the mean of time taken for the 100 ±0.1 ml of 0.1% methylene blue
solution to decolourise in 5 bottles in 3 trials for point A, B, and C in River A and
B.
The mean is calculated by using this formula:
Example of calculating mean in river A at point A in trial 1 :
𝑥̅ = ∑ 𝑥 /𝑛
=
85+86+84+85+86
5
= 85.2
Refer the appendix 4 for the calculated mean of 90 water samples of river A and river B.
7.2 Calculating the overall mean of time taken for the 100 ±0.1 ml of 0.1%
methylene blue solution to decolourise of 3 trials for Point A, B and C in River A
and B.
Example of calculating mean of 3 trials for point A in River A :
𝑥̅ = ∑ 𝑥 /𝑛
=
85.2+85.2+89.6
3
= 86.7
Refer appendix 5 for the calculated overall mean of time taken for the 100 ±0.1 ml of 0.1%
methylene blue solution to decolourise for 3 trials for Point A, B and C in River A and B.
𝑥̅ = ∑ 𝑥 /𝑛
where :
𝑥̅ is the mean
∑ stands for ‘sumof’
𝑥 refers to theindividualvalues ofthesample
𝑛 is the total numberofindividualvalues in the sample.

15. 15
7.3 Calculating standard deviation of time taken for the 100 ±0.1 ml of 0.1%
methylene blue solution to decolourise in 3 trials for point A,B and C in River A
and B.
The standard deviation, S is calculated by using this formula :
Example of calculating standard deviation of time taken for the methylene blue solution to
decolourise in trials 1 for point A in River A :
𝑆 = √∑ 𝑥2−
(∑ 𝑥)
2
𝑛
𝑛−1
= √(852 +862 + 842 + 852 + 862 )−
(85+86+84 +85+86)2
5
4
= 0.8366
= 0.8
Refer appendix 6 for the calculated standard deviation of time taken for the 100 ±0.1 ml of
0.1% methylene blue solution to decolourise in 3 trials for point A,B and C in river A and B.
𝑆 =
√∑ 𝑥2 −
(∑ 𝑥)
2
𝑛
𝑛 − 1
where :
∑ stands for ‘sum of’
x refers to the individual valuesof the
sample
n is the total number of individual
values in the sample

16. 16
7.4 Calculating the average of standard deviation of time taken for the 100 ±0.1 ml
of 0.1% methylene blue solution to colourless for point A, B, and C in river A
and B.
Example of calculation the average of standard deviation of time taken for the methylene blue
solution to colourless for point A in River A :
𝑥̅ = ∑ 𝑥 /𝑛
=
0.8+0.8+0.5
3
= 0.7
Refer appendix 7 for the calculated average of standard deviation of time taken for the 100
±0.1 ml of 0.1% methylene blue solution to colourless for point A, B, and C in river A and
B.
7.5 Calculating t-test
i. The value of t is calculated by using this formula:
t =
𝑥̅ 𝑎− 𝒙̅ 𝒃
√[
( 𝑛 𝑎−1 ) 𝑆 𝑎
2
+(𝑛 𝑏−1 )𝑆 𝑏
2
𝑛 𝑎+ 𝑛 𝑏−2
][
𝑛 𝑎+ 𝑛 𝑏
𝑛 𝑎 . 𝑛 𝑏
]
where :
𝑥̅ 𝑎 is meanforriverA
𝑥̅ 𝑏 is meanforriverB
𝑛 𝑎 is the total numbersamplesforriverA
𝑛 𝑏 is the total numbersamplesforriverB
𝑆 𝑎 is the standarddeviationforriverA
𝑆 𝑏 is the standarddeviationforriverB

17. 17
Example of calculation of t value :
t =
88.1−65.7
√[
(44)(0.72)+(44)(8.072 )
88
][
90
2025
]
=
22.4
√(32.80745)(
2
45
)
= 18.55038979
= 18.6
ii. The value of degree of freedom, d.f is calculated by using this formula :
Example of calculation of degree of freedom :
d.f = 45 + 45 -2
= 88
d.f = 𝑛 𝑎 + 𝑛 𝑏 − 2
where :
𝑛 𝑎 is the total numberof samplesof riverA
𝑛 𝑏 is the total numberof samplesof riverB

18. 18
7.6 Presentation of analysed data
Figure 9 shows the comparison of the time taken for the 100 ±0.1 ml of 0.1% methylene blue
solution to colourless on river A and river B.
0
5
10
15
20
25
< 50 50 - 54 55 - 59 60 - 64 65 - 69 70 - 74 75 - 79 80 - 84 85 - 89 90-94
Frequancy,%
Class/ min (±0.1)
River A
River B

19. 19
8.0 CONCLUSION AND EVALUATION
Since the calculated t value is greater than the tabulated value of t value, the null hypothesis
can be rejected (refer to appendix 8).The calculated t value is 18.6 while the tabulated t value
is 1.99. A significance different at p = 0.025 means that we have 95% confidence level if the
null hypothesis is accepted (there is no difference between two mean). So that we can be
reasonably confident that the samples do not differ from each other. But, we still have 5%
nearly to be wrong in this statement in reaching conclusion. The time taken for the 100 ±0.1
ml of 0.1% methylene blue solution to decolourise in river B is shorter than river A as the
water contains low dissolve oxygen in river B. Also, river B has cloudy (green – brownish) in
colour, oily structures on the surface of the river, present of algae, have no fish and have
unpleasant smell. This may be due to the larger drowning of sewage and fertilizers into the
river and so it enhances the growth of algae on the surface of water. This shows that the river
B is more polluted compared to river A as the BOD in river B is higher than river A.
Now we compare the calculated t value with the tabulated t value for higher levels of
significant, p=0.01 (refer appendix 9). The calculated t value is 18.6 and the tabulated t value
is 2.37. Since the calculated t value is higher than the tabulated t value, the null hypothesis
can be rejected. At this higher levels of significance (p=0.01), enables to tell us that the
probability of our conclusion being correct. There is 98% of means being highly significant.
Hence, by using this statistical test (t test), it only allow us to make a statement with a degree
of precision but not allow us to prove or disprove anything (Student's t-test). With 95%
confidence are good enough to support a conclusion with a significant result at the 95%
probability levels (Student's t-test). The larger the value of t, the larger the difference between
the means of the two samples (Morgan, 2008), the more certain that the two sets of data are
different.
Based on the bar graph, it shows the clear comparison of the time taken for the 100 ±0.1 ml
of 0.1% methylene blue solution to decolourise on river A and river B. For the interval of 80-
84 minutes time taken for the 100 ±0.1 ml of 0.1% methylene blue solution takes to
decolourise, the frequency of river A is higher than river B which is water samples in river A
is 2.2% of water samples have longer time taken for the 100 ±0.1 ml of 0.1% methylene blue
solution takes to decolourise. Also from the interval of 85-89 minutes, the frequency of water
samples of river A which is 24.4% of water samples being higher time taken compared to
river B which is only 4.4% of water samples that takes 85 to 89 minutes for the 100 ±0.1 ml

20. 20
of 0.1% methylene blue solution takes to decolourise. For the intervals 90-94 minutes, time
taken for the 100 ±0.1 ml of 0.1% methylene blue solution to decolourise, its only have
involve water samples of river A which is 21.2% of water samples. This shows that river A
takes longer time taken to decolourise the 100 ±0.1 ml of 0.1% methylene blue solution
compared to river B.
In a nutshell, the main factor for this pollution is not the weedy macrophytes, overgrowths of
periphytons, and blooming cyanobacteria but the activity of human itself. People have only
themselves to blame when lakes become unsuitable for swimming, lose their good fishing
qualities, or poison animals that drink from them. Much research has shown that harmful
cyanobacterial growths result from phosphate pollution that is linked to human activities
(E.Graham, M.Graham, & W.Wilcox, 2003).

21. 21
9.0 LIMITATIONS AND WAYS TO IMPROVE
At the first stages of experiment, I used to test the water samples by placing it in the marble
pallet. The water samples that added with the 100 ±0.1 ml of 0.1% methylene blue solution is
observed for three days and there is no colour changes occur.
Figure 10 shows the colour of the water samples that is tested with 100 ±0.1 ml of 0.1%
methylene blue solution for three days and does not shows any colour changes.
Unfortunately, in order to run this experiment, the solution(water samples and the 100 ±0.1
ml of 0.1% methylene blue solution need to be in the closed area which then I use the reagent
bottles with stoppers to prevent dissolve of atmospheric oxygen.
The use of the six stopwatches made it easier to read the time taken for the methylene blue
solution to colourless. However, cheap stopwatches are not particularly accurate and so the
reading of time may have been out by a second. I am far more confident with the readings
obtained using the digital stopwatch, which gave me more precision. Even all of bottle
samples were located in the dark cupboard, but some of the bottle samples have been
penetrate by slightly by light. However, I still believe that the reading of the time taken that I
take were as accurate as I could make them in circumstance.
Sampling was systematic, but in a few cases, I unable to take the samples from the centre of
the river due to the depth of the river and the torrents flow of water. Some point on the river
has torrent flow of water and some point have slow flow of water. Taking water samples
from the centre of the river were the best technique used in water sampling.

22. 22
Dividing the river into three points was done on the basis of length of 100 m between each
point. However, in practice it was more difficult to take the samples from the centre of the
river due to the steep canyon. To some point, I unable to take the water samples by using the
same technique and have to modify the technique that I used to do such as by using
innovation of the bail.
Figure 11 shows the modified the scoop with the long wooden stick.
Some of the samples did produce outlier’s results, which was unexpected. Larger sample
sizes may have overcome this problem.
Abstract
One of the objectives in my research is to alert people about the consequences of water
pollutions. Water pollution can occur by several factors. By doing this research, I’m
comparing the two factors that lead to water pollution which are commercials fertilizers and
sewage drowning such as domestic waste to see whether they are really can affect the
cleanliness of the water. Thus, this research will expose the community about the factor that
can contribute to the water pollution.
By using manual grab sampling technique, I’m collecting 90 water samples in two different
rivers that have different source (45 samples for each river) and test the water samples with
100 ± 0.1 ml of 0.1% methylene blue solutions. Then, the time taken for the methylene blue
solution to decolourise is recorded.
River B shows the shortest time taken for the 100 ± 0.1 ml of 0.1% methylene blue solution
to decolourise (with the mean of 65.7 ± 0.1 min) while River A takes an average time of 88.1

23. 23
± 0.1 min for the 100 ± 0.1 ml of 0.1% methylene blue solution to decolourise. A bar graph is
constructed to make a comparison between river A and river B. In the interval of 60-64 min
(short time taken), river B shows concentrated data in this time interval. A statistical test is
conducted which is t test to make a proven statement about the null hypothesis. The
calculated t value for the two mean of the two rivers (t = 18.6) is greater than the tabulated
value (1.99). Since the calculated t value is higher than the tabulated value, the null
hypothesis can be rejected.