Exercise Lab 6 Graphing and Data_Excel Steps-1
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Lab 6: Graphing and Data Analysis
EXERCISE
Exercise 1: Analyzing the Data
For today’s lab, you will examine data from actual experiments conducted in the Fisher lab here at UNC. The hypothesis for the experiment is that the older life cycle stages will be more resistant to the effects of the pesticide cypermethrin. This pesticide is commonly used in agriculture to kill a wide variety of insect (arthropod) pests. However, it is a non-specific pesticide and, therefore, kills many different arthropods including copepods. Although it is a pesticide approved for use on land, much of it can run off into the oceans, and there has been very little testing to determine its effects on aquatic species. Therefore, this study was conducted to look at these effects, especially on the younger, and therefore potentially more vulnerable, life cycle stages. The life cycle stages examined were the naupliar and adult stages. For the experiment, nauplii were placed individually into wells of a 24 well plate. This means that each plate had 24 nauplii and each Nauplius was placed in its own well. The wells were filled with a mixture of seawater and cypermethrin, and the survival of each Nauplius was recorded after 48 hours. This was repeated for each life cycle stage, and then the entire experiment was repeated three times. The raw data
are shown below (0 = dead, 1 = live):
Naupliu
s
Experiment 1
Experiment 2
Experiment 3
Experiment 4
Well #
Alive or Dead
Alive or Dead
Alive or Dead
Alive or Dead
1
1
1
0
0
2
0
0
0
1
3
1
0
0
0
4
1
1
1
0
5
0
0
1
0
6
0
0
1
1
7
0
0
1
0
8
1
0
0
0
9
0
1
0
1
10
0
1
1
0
11
0
0
0
1
12
1
0
1
0
13
1
0
0
0
14
1
0
0
0
15
0
1
0
1
16
0
1
1
0
17
0
1
0
0
18
0
0
0
0
19
0
0
0
0
20
0
0
1
1
21
0
0
0
0
22
1
1
0
0
23
0
1
0
0
24
1
0
0
1
Adult
Experiment 1
Experiment 2
Experiment 3
Experiment 4
Well #
Alive or Dead
Alive or Dead
Alive or Dead
Alive or Dead
1
1
0
0
0
2
1
1
0
1
3
1
1
1
1
4
0
0
1
1
5
0
0
1
1
6
0
1
1
0
7
1
1
1
1
8
1
1
1
1
9
1
1
0
1
10
1
0
1
1
11
1
1
1
0
12
0
1
1
1
13
1
1
0
1
14
1
0
1
1
15
0
1
0
1
16
1
1
1
0
17
1
1
1
0
18
0
1
0
1
19
1
0
0
1
20
1
1
1
0
21
1
0
1
0
22
0
1
1
1
23
1
1
1
1
24
0
1
1
1
*There are a couple more things to consider before you process any of the data. The first is copy errors
. You could go line by line for each of the above tables and enter them into Excel by hand. However, you are a human being, and we all make mistakes. Even if you only make a mistake in entering one out of a hundred boxes, there are 288 data points above so that means that 2 or 3 of the above data points would be wrong. That doesn’t sound like a lot, but it could be enough to change the results of your statistical analysis. If your error rate were only to go up to three out of a hundred than 8 or 9 of them would be recorded wrong. So whenever possible use a computer to copy raw data points like that, Excel will get that data copied correctly 100% of the time.
The second thing to consider is to use your RAW data for all your calculations. We’ve discussed the importance of large sample sizes. If you were to calculate the proportions, averages or standard error of each experiment (like you will below) then run a t-test on those instead of your raw data; then you just reduced your sample size from 96 for each group to 4. That kind of systemic error virtually ensures that your results will not be statistically significant.
Finally, use Excel to perform all these calculations. You could use a calculator for some of this but then you’re entering data by hand (remember copy errors?). In addition, Excel is a very common program that you will use
over and over again. You may not know all the ins and outs of it right now, but the more contact you have with it the sooner you will. It’s like when you first started using Word or any other program; until you had a certain amount of experience with it some things were difficult. When you get used to using Excel these things
will be much easier, make it do all the hard work for you!
Step 1: Enter your data into an Excel File/Google Sheets
*This can be done by hand OR you can download the word document, highlight the table, and copy and paste it into Excel. Do this for the adult copepod data and the nauplii data. I would recommend putting each table on two different “Sheets” To make a new “sheet” in Excel, look at the bottom left where it says “Sheet 1” There is a + next to that. Click on the + to make a new sheet.
Step 2:
Calculate the proportion (or fraction) of organisms that survive in each experiment. This can also be thought of as the Average rate of survival.
If you were to do this by hand you would add up all of the numbers in the column (1’s and 0’s) and then
divide by the total number of wells in a column (in this case 24) to give you your average. Excel makes it easier for us! In an empty cell below a column, enter the equation =AVERAGE(highlight the column you want to find the average of)
. This does the same thing. Repeat this equation at the bottom of each column.
*Caution – these cells say L3:L26 – yours will be different depending on where you put your table in Excel. Also, make sure you close the parentheses.
You should now have an average or proportion for each column of data
.
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If you haven’t done so already create two tables in Excel for proportion survival of Naup. and proportion survival of Adults. It may look something like this. Using the function we just learned of =AVERAGE(), now calculate the average of the averages, or the overall average for the Nauplii survival
and the overall average for Adult survival
. Step 3. Calculate averages and standard error for each age group
N = your sample size
In ecology, each well plate
is considered as 1 sample
, even though there are multiple wells. This is because they are all in the same environment and so considered as one. If each well plate is 1 sample, what is the total
“n” for Nauplii? For adults?
To calculate standard error in excel: There is no formula for standard error in excel, so you must create it. Essentially, you use the standard deviation formula and then divide this by the square root of the sample size (n).
a. In a cell below your data, type =(STDEV(B2:E2))/SQRT(n) and hit Enter. (*Your N=the number you found earlier for the sample size.) *** Your cell numbers may be different than B2:E2 if you have your data in different cell.
b. CAUTION: The equation above will work if your data are in cells B2 to E2. Be sure to change the cells to fit where you typed your data!!
c. Do this procedure for both Nauplii data and then Adult data. Remember to label your answers so you know
which is which
!
d. Now create a table with your Overall average and your Standard error
Nauplius
Adult
Average
Standard Error
Step 4: Conduct a t-test to compare each life cycle stage to the other. To do this, go to the following link http://www.graphpad.com/quickcalcs/ttest1.cfm
. Select mean, SEM, and n. The mean is your overall average, the SEM is your standard error, and “n” is your sample size that you calculated earlier. (Remember one well plate is one sample, not each individual well.)
Enter your data, then click on Enter Now.
Record your T-statistic and p-value.
Nauplius vs. adult
T-statistic
38.5524
p-
value
46
Significant difference (yes or no)
0.009
Does your t-test indicate a significant difference between the two life cycle stages?
Yes, the t-test indicates a significant difference between the Nauplius and adult life cycle stages.
How do you know the difference is (or is not) significant?
The p-value associated with the t-test is 0.009. Since this p-value is less than the conventional significance level of 0.05. Therefore, there is evidence to suggest a significant difference between the two life cycle stages.
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Putting it all together now:
First, restate the original hypothesis for this experiment.
Null Hypothesis (H0): There is NO statistically significant difference between the Nauplius and adult life cycle stages.
Alternative Hypothesis (H1): There IS a statistically significant difference between the Nauplius and adult life cycle stages.
Now, do the results of the t-test support your hypothesis? If yes, why do you think this is so? If not, why not?
Yes, the results of the t-test support the hypothesis. The p-value is below the significance level, indicating that the observed difference between Nauplius and adult life cycle stages is unlikely to
have occurred by random chance. Therefore, we reject the null hypothesis in favor of the alternative hypothesis.
When conducting research, you can have a t-test that shows a significant difference and still NOT support your
hypothesis? Explain how this could occur.
Yes, even if the t-test shows a statistically significant difference, it may not support the hypothesis if the observed difference is not practically significant or relevant. Practical significance considers whether the difference, although statistically significant, is meaningful in the real-world context. Additionally, issues such as
sample size, study design, or confounding variables can affect the interpretation of results.
Step 4: Present your data in graphical form. (Hint: Look over the introduction for this lab to decide what type of graph might best represent your data.) You can also check out these links:
https://towardsdatascience.com/data-visualization-101-how-to-choose-a-chart-type-9b8830e558d6
https://www.skillsyouneed.com/num/graphs-charts.html#:~:text=There%20are%20several%20different
%20types,pie%20charts%2C%20and%20Cartesian%20graphs
.
1
2
3
4
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Chart Title
Nauplius
Adult
Exercise 2
Sam is examining the effects of gasoline on copepod survival. He hypothesizes that increased levels of gasoline will lead to high rates of mortality. He has a control group and two experimental groups. Sam conducts 6 trials (6 well-plates) for
each group. He then calculates the individual proportions for each trial.
Use his data to answer the following:
1. What is the average proportion for the control group? For the 0.01 group? For the 0.001 group?
Control Gr
0,886666667
0.01
0,77
0.001
0,68
2. Looking at the averages, can you tell if there is a significant difference between the groups? Why or why not?
there is a significant difference, and the decreasing trend aligns with the statistical findings.
Sam’s Data
Proportion of Copepods living after 1 week
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Trial #
Control (No gasoline)
0.01 ppt gasoline
0.001 ppt gasoline
1
0.95
0.83
0.73
2
0.97
0.79
0.63
3
0.83
0.71
0.71
4
0.85
0.85
0.53
5
0.79
0.69
0.69
6
0.93
0.77
0.78
3. Now calculate the standard error for each group. (You will need to refer to STEP 3 from the first exercise)
What is the standard err for the control group? For 0.01? For 0.001
Control Group
0,029851484
0.01
0,026034166
0.001
0,035815887
4. Using the t-test website, compare the control with each of the test groups. link http://www.graphpad.com/quickcalcs/ttest1.cfm
Record the p-value and t-statistic for each comparison. Is it significant? How do you know?
Control vs 0.01
Control vs 0.001
p-value:10
p-value: 10
t-statistic:
2.9732
t-statistic:
0.6939
Significant?
0.040
Significant?
0.303
Exercise 3
Shani is investigating how often meadowlarks sing in urban (city) vs rural (country) environments. She hypothesizes that urban birds will sing more often than rural birds because they will need to communicate more frequently.
Shani puts a summary of her data in a table. Are her results statistically significant? How do you know?
Is Shani’s hypothesis correct? How do you know?
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- Using satellite imagery to aid agriculture grain production has been done on a limited scale as far back as the 1970's when NASA first launched the Landsat satellite. But it has only been recently that the satellite images, including GPS satellites, and data from typical farm operations have been integrated together to provide a data driven innovative technology that improves the efficiency and productivity on farms. In one study samples were taken on wheat yields in the State of Washington which historically had average yields of 35.6 bushels per acre with a standard deviation of 3.7 bushels per acre. Suppose that a random sample of the wheat yields on 40 different farms was taken and the sample mean for the wheat yields was computed, and you are interested in a probability about the sample mean. 28. Pick the equation most appropriate to deal with the problem above. a) p(x) = (nCx) p*(1-P)"-* b) x = μ + zo c) z= x-μ of √n d) z = x-μ O 29. If the sample mean is 34.8 bushels of wheat,…arrow_forwardThe stated goal of a chain of service stores is to provide a lube and oil change for anyone's automobile in 15 minutes. The manager at one store thinks that there is a growing disparity among his workers in the time it takes to lube and change the oil of an automobile. To monitor this, the manager has selected a sample of 20 days and has recorded the time it took to service 100 automobiles each day. Data about the number of times the service was performed in 15 minutes or less (515) are found below. Complete parts a through c Click the icon to view the data. a. Compute p and sp p-(Round to four decimal places as needed.) Sp (Round to four decimal places as needed) b. Construct a p-chart and determine if the process of the time required for oil and lube jobs is in control Which graph below shows a p-chart for the data? O A. OB. Am 0.05 -ww Is the process in control? OA. Since 10 points in a row are below the centerline, followed by 10 points above, conclude that the process is out of…arrow_forwardTwo of the hottest smartphones on the market are the newly released iPhone6 and the Samsung Galaxy S6. CNet.com offers online reviews of all major cell phones, including battery life tests. In a review of the iPhone6, the talk-time battery life of 35 iPhones was measured. Similarly, the talk-time battery life of 30 Galaxy S6s was measured. Two outputs are given below. Which is appropriate for analyzing the data collected? Output 1 Output 2 Using the StatCrunch output chosen above, determine if there is a difference in the mean battery life for the two phones. Use a significance level of 0.01 when conducting the test. Select the appropriate hypotheses. Make sure the notation used in the hypotheses agrees with the type of samples selected in the output. Ho:μd=0Ho:μd=0Ha:μd<0Ha:μd<0 Ho:μd=0Ho:μd=0Ha:μd≠0Ha:μd≠0 Ho:μd=0Ho:μd=0Ha:μd>0Ha:μd>0 Ho:μ1=μ2Ho:μ1=μ2Ha:μ1>μ2Ha:μ1>μ2 Ho:μ1=μ2Ho:μ1=μ2Ha:μ1≠μ2Ha:μ1≠μ2 Ho:μ1=μ2Ho:μ1=μ2Ha:μ1<μ2Ha:μ1<μ2 αα =…arrow_forward
- Variation in gasoline mileage among makes and models of automobiles is influenced substantially by the weight and horsepower of the vehicles. An automobile manufacturer would like to determine if heavier cars consumer more gas than lighter cars. He collects data on 25 heavy and 25 light cars to test this hypothesis. What analysis would you complete to answer this question?arrow_forwardIn an observational study, it was noticed that people that eat more chocolate tend to have better blood flow than the general population. Researchers wonder whether the improved blood flow might be caused by flavanols (a molecule found in many foods). Dark chocolate contains a high concentration of flavanols, but milk chocolate contains a much lower concentration. How could the researchers design an experiment to determine the effects of flavanol on blood flow from the different types of chocolate?arrow_forwarduse calculator methodarrow_forward
- People gain weight when they take in more energy from food than they expend. Researchers wanted to investigate the link between obesity and energy spent on daily activity. Choose 20 healthy volunteers who don't exercise. Deliberately choose 10 who are lean and 10 who are mildly obese but still healthy. Attach sensors that monitor the subjects' every move for 10 days. The table below presents data on the time (in minutes per day) that the subjects spent standing or walking, sitting, and lying down. Is there a significant difference between the mean times the two groups spend lying down? Let ?1 be the mean time spent lying down by the lean group, and ?2 be the mean time for the obese group. Time (minutes per day) spent in three different postures by leanand obese subjects Group Subject Stand/Walk Sit Lie Lean 1 506.100 374.300 559.500 Lean 2 602.925 376.512 455.650 Lean 3 315.212 587.138 536.362 Lean 4 587.644 352.144 487.269 Lean 5…arrow_forwardPeople gain weight when they take in more energy from food than they expend. Researchers wanted to investigate the link between obesity and energy spent on daily activity. Choose 20 healthy volunteers who don't exercise. Deliberately choose 10 who are lean and 10 who are mildly obese but still healthy. Attach sensors that monitor the subjects' every move for 10 days. The table below presents data on the time (in minutes per day) that the subjects spent standing or walking, sitting, and lying down. Is there a significant difference between the mean times the two groups spend lying down? Let ?1 be the mean time spent lying down by the lean group, and ?2 be the mean time for the obese group. Time (minutes per day) spent in three different postures by leanand obese subjects Group Subject Stand/Walk Sit Lie Lean 1 509.100 366.300 554.500 Lean 2 603.925 374.512 452.650 Lean 3 322.212 580.138 533.362 Lean 4 581.644 362.144 489.269 Lean 5…arrow_forwardZika and male reproduction. The Zika virus, which is known to be transmitted by mosquitoes, can also be transmitted sexually, as the virus can persist for weeks in the semen of symptom-free infected men. Using a mouse model, researchers examined the impact of Zika on male reproductive organs. Young adult male lab mice of a breed susceptible to Zika were either inoculated with Zika or not. Here are the weights of their testes two weeks later (in grams, g) Uninfected 85 85 90 91 96 100 Infected 17 93 15 12 10 9 10 12 a) What two populations do the researchers want to compare and what is the response variable? b) State the null and alternative hypotheses. Plot the data and check the conditions for inference. Explain why a two-sample t test is not appropriate here.arrow_forward
- answer part b frq stylearrow_forwardA professor suspects that recent testing will lead to greater retention of course material in a statistics course.to examine this, students in one session of PSYC stats are given a short quiz every week. Students in the second section of the PSYC stats take only one mid term and final. At the end of the quarter, both sections receive the final exam, and the scores are summarized below. Does the data indicate more frequent testing will produce and effect on the final exam score? Use a two-tail test at .05 level of significance. Show your work for each step Frequent exams N= 22 Sample means = 85 Ss=350 Frequent exams N= 20 Sample means = 79 Ss=1050 Calculate Cohens d or argue why its not appropriate in this scenario? Name the research design used? Write the results in APA formatarrow_forwardA medical student at a community college in city Q wants to study the factors affecting the systolic blood pressure of a person (Y). Generally, the systolic blood pressure depends on the BMI of a person (B) and the age of the person A. She wants to test whether or not the BMI has a significant effect on the systolic blood pressure, keeping the age of the person constant. For her study, she collects a random sample of 150 patients from the city and estimates the following regression function: Y= 15.50 +0.90B + 1.10A. (0.48) (0.35) The test statistic of the study the student wants to conduct (Ho: B, =0 vs. H4: B, #0), keeping other variables constant is. (Round your answer to two decimal places.) At the 5% significance level, the student will v the null hypothesis. Keeping BMI constant, she now wants test whether the age of a person (A) has no significant effect or a positive effect on the person's systolic blood pressure. So, the test statistic associated with the one-sided test the…arrow_forward
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