The Experiments with the Amount of Oxygen Found in Water - Lab Report Example

Biology 23rd September Exercise Data interpretation As the level of dissolved Oxygen increases from 0 to 12 the number of fish in the water rises. Beyond 12 the population of fish falls drastically and then gradually the population rises again. 2. The greater the amount of dissolved Oxygen in water the greater is the number of fish observed. 3. Water temperature, weather and pollution also contribute to the amount of oxygen found in water. Basically, the experiment can be divided into two parts. First part comprises of taking clean 10 or 11 (depending on the number of observations one wants) small fish tanks of equal sizes and filling them with normal same tap water. Once done, the fish tanks must be kept in a room with same weather, level of pollution and temperature. Temperature can be controlled using an air conditioner. It is important to control such factors because the amount of Oxygen in water can vary if water temperature, weather and pollution vary. Once, this has been done the time has come to introduce varying levels of Oxygen into the water in fish tanks. This can be done by using various ways. On simple way is to use an Oxygen generator, air regulator, plastic tubing and diffusion stone connect these together and pump Oxygen into water for a specific time. Since one is working in a lab equipment must be chosen according to lab’s size and equipment available. The concentration of Oxygen in water can be adjusted initially (starting from 6.5mg/L) by changing the duration of the pumping. Once, Oxygen has been pumped into water its concentration can be measured and recorded using an LDO Sensor (hack). The water in all 10 fish tanks must contain different amounts of Oxygen. Now, fishes of the same species and same age must be introduced into the fish tanks. For accurate results it is best to select fishes that can survive at a mediocre concentration of Oxygen (6.5 mg/L) and room temperature (25 degrees Celsius). An example of such fishes is Smallmouth bass. Same number of fishes must be introduced into all of the fish tanks. Every day fishes must be given the same brand of fish food and the same amount of food (To ensure this before food is given it can be weighed using digital balance). After a week, the number of fish left alive in the tank can be counted by removing them using a net and counting manually. 4. Independent variable is the amount of Oxygen dissolved in water while dependent variable is the number of fish observed. 5. The control in this experiment will be a fish tank filled with Smallmouth bass fishes. The quantity of fishes will be same as in all other cases. The fish tank will be placed in the lab where temperature will be 25 degrees Celsius. Temperature will be kept constant using an air conditioner. The fishes will be given same amount and kind of food as fishes in other tanks. However, the concentration of Oxygen dissolved in water will be 0 mg/L. To remove all of the Oxygen from water it can be heated in a sauce pan while recording the temperature using a Candy thermometer. When temperature reaches 286 Fahrenheit Oxygen leaves from the water. 6. A scatter plot would be appropriate for the given data set. Such a plot would help in clearly showing if there is any kind of correlation between the two variables. 7. Correlation coefficient: 0.822458 8. By looking at the scatter plot and the correlation coefficient calculated one can conclude that amount of Oxygen dissolved in water is strongly correlated with the number of fish observed in the water. As one variable increases the other variable is highly likely to increase. Exercise 2: 1. Observation is quantitative. Null hypothesis: Plant placed on window sill does not grow 3 inches faster per day compared to plant on coffee table in the middle of living room. Hypothesis: If Plant placed on window sill then it does grow 3 inches faster per day compared to plant on coffee table in the middle of living room. Experimental approach & data collection: I will take two pots of same size and fill them up with the same amount of soil. After this I will plant two plants of the same height and species into the pot. One pot will be put on window sill and another pot will be placed in the middle of the living room. At the end of every day heights of both plants will be measured using a tape and compared. The difference will be recorded for a week. Finally, an average of the difference will be found to know the average comparative growth rate of the plant on the window sill. Dependent variable: Comparative growth rate of place placed on window sill. Independent variable: Whether plant is placed on window sill or coffee table in living room. Data presentation: Day number Plant kept on windowsill’s height (plant A). Plant kept in middle of the living room height (plant B). Difference in height= Comparative growth rate of plant A/day. Sum of difference in heights Sum of days Average comparative growth rate of plant A/day Data Analysis: Compare Average comparative growth rate of plant A/day is with comparative growth rate of 3 inches per day. Negative control: Take the plant in pot and place it in a dark cupboard with minimal ventilation. Positive control: Take the plant and put it outside in the garden where sunlight falls on the plant. 2. Not testable. 3. Quantitative. Null hypothesis: If Sally exercises regularly and eats healthy foods there is no change in her blood pressure. Hypothesis: If Sally exercises regularly while eating healthy foods there is a decrease in her blood pressure by 10 points. Experimental approach & data collection: Put Sally on a diet comprising of healthy foods only (diet should be recommended after consulting a trained nutritionist only) and hire a trainer who ensures that she exercises daily for 30 minutes (as recommended by doctors) while following her diet. At the end of each day Sally’s blood pressure will be measured using a digital meter and recorded. Once, data has been collected for a month it will be analyzed. Dependent variable: Sally’s blood pressure. Independent variable: The strictness with which Trainer insures Sally follows her diet and exercises. Data presentation: Blood Pressure (mm/Hg) Start of the month Last day of the month Change in Blood pressure (Starting BP- Ending BP) Data Analysis: Compare change in blood pressure during the month with what the hypothesis states. If the decreases are same then accept the hypothesis. Negative control: Allow Sally or a woman of her age, BMI and race to follow the same diet and exercise routine Sally followed during her ordinary non-experimental days. Positive control: Select a woman of Sally’s age, BMI and race. Put her on the healthiest of diets and exercise routine. Her blood pressure is bound to decline rapidly. 4. Not testable. 5. Not testable. 6. Not testable. Exercise3: Conversion For each of the following, convert each value into the designated units. 1. 46,756,790mg=46.75679kg 2. 5.6 hours=20160 seconds 3. 13.5cm=5.32 inches 4. 47°C=116.6°F Exercise 4: 1. Average push-ups in a minute=66.25 %error= (45-66.25)/45*100=47.2% (2 is recurring) Deviation = (average - actual) Actual Deviation 64 2.25 69 -2.75 65 1.25 67 0.75 Sum of deviations 1.5 Standard deviation=Sum of deviations/number of measurements=1.5/4=0.375 The experimental results are precise as shown by the low standard deviation; however, the percentage error is very high indicating that the results are inaccurate. 2. The average score for the 5th grade math test is 89.5. The top 4th graders took the test and scored 89, 93, 91 and 87. Average score in a test by 4th grader=90 %error= (89.5-90)/89.5*100=0.559% Deviation = (average - actual) Actual Deviation 89 1 93 -3 91 -1 87 3 Sum of deviations 0 Standard deviation=Sum of deviations/number of measurements=0/4=infinite The experimental results are not precise as shown by the standard deviation; however, the percentage error is very low indicating that the results are accurate. 3. Average temperature=89 %error= (75-89)/75*100=18.68% Deviation = (average - actual) Actual Deviation 89 0 88 1 90 -1 Sum of deviations 0 Standard deviation=Sum of deviations/number of measurements=0/3=infinite The experimental results are not precise as shown by the standard deviation and the percentage error is noticeably high indicating that the results are inaccurate. 4. Average score=(3+1+1+1)=6/4=1.5 %error= (3-1.5)/3*100=50% It is not accurate due to very high percentage error Sum of deviations=-1.5+0.5+0.5+0.5=0 Standard deviation=Infinite, therefore precision is nonexistent. 5. A local grocery store was holding a contest to see who could most closely guess the number of pennies that they had inside a large jar. The first six people guessed the numbers 735, 209, 390, 300, 1005 and 689.The grocery clerk said the jar actually contains 568 pennies. Average guess=554.67 %error= (568-554.67)/568*100=2.35% Deviation = (average - actual) Actual Deviation 735 -180.33 209 345.67 390 164.67 300 254.67 1005 -450.33 689 -134.33 Sum of deviations 0.02 Standard deviation=Sum of deviations/number of measurements=0.02/6=3.3x10^-3 The experimental results are precise as shown by the standard deviation and the percentage error is very low indicating that the results are accurate. Exercise 5: Part 1: Determine the number of significant digits in each number and write out the specific significant digits. Questions Number of significant digits specific significant digits 1. 405000 3 4,0,5 2. 0.0098 2 9,8 3. 39.999999 8 3,9 4. 13.00 4 1,3 and both zeroes 5. 80,000,089 8 8,9, all zeroes 6. 55,430.00 7 5,5,4,3,0(all zeroes) 7. 0.000033 2 Both 3s 8. 620.03080 8 All Part 2: Write the numbers below in scientific notation, incorporating what you know about significant digits. 1. 70,000,000,000=7x10^10 2. 0.000000048=48x10^-9 3. 67,890,000=6789x10^4 4. 70,500=7.05x10^4 5. 450,900,800=4.509008x10^8 6. 0.009045=9.045x10^-3 7. 0.023=2.3x10^-2 Bibliography Hack, Michael. "Optical measurement of concetration of Oxygen in water." n.d. www.hach-lange.com. HACH Lange. 22 September 2012 . Read More