ENV 250 Module Five Homework

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Southern New Hampshire University *

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250-X6

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Biology

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Nov 24, 2024

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docx

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ENV 250 Module Five Homework Instructions: In this assignment, you will draw conclusions regarding the significance of findings, determine if findings are biologically important, and determine the difference between accuracy and precision as well as the difference between reproducibility and replication. For each problem below, please address the prompt and explain your reasoning. 1. Report the mean (including standard deviation), median, and mode for the following data set of worm lengths measured in centimeters: 6, 4, 11, 12, 13, 9, 10, 14, 8, 4, 7, 12. Be sure to show your work, including the formula you used to calculate the mean, median, and mode, and use the proper units. 4,4,6,7,8,9,10,11,12,12,13,14 Mean=9.166666667=9.2 cm Mode=4cm,12cm Median=9.5cm For the mean I added all the numbers in data set and divided by 12 because that is the amount of data entries. The mode is found by ordering the numbers and looking at how many times they appear in the set the number appearing most is the mode if you have 2 frequent numbers appearing they are both a mode. The median is found by finding the middle number but if there is a even amount of points you must take both middle numbers add them together and divide by 2. Standard deviation is found by finding the mean, then for each data point find the square of its distance to the mean. s=√1n−1∑ni=1(xi−¯x)2 s = 1 n − 1 ∑ i = 1 n ( x i − x ¯ ) s= 3.4067668846078 (4-9.1666666667) 2 +….+(14-9.1666666667) 2 /12-1=127.6666666667/11=11.6060606061=sqrt11.6060606061=3.4067668846078 cm. 2. The Department of Fish and Wildlife looked at the ages of individuals between 65 and 81 that frequented the Bolsa Chica Reserve during the week of June 13 th , 2021 to help determine if more facilities and resources are necessary to support the needs of that age group. They determined there were 40 individual visitors that fit those criteria, with the following ages: 65, 66, 66, 67, 67, 67, 68, 68, 68, 68, 69, 69, 69, 69, 69, 69, 70, 70, 70, 70, 70, 71, 71, 71, 71, 72, 72, 72, 73, 73, 73, 74, 74, 75, 75, 76, 77, 78, 79, and 80. Age (years) Number of Visitors 65 1 66 2 67 3 68 4 69 6 70 5 71 4 72 3 73 3 74 2 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81+ 0 1 2 3 4 5 6 7 Number of Visitors to Bolsa Chica Reserve (06/13/2021 - 06/19/2021 Age (years) Number of Visitors
Age (years) Number of Visitors 75 2 76 1 77 1 78 1 79 1 80 1 81+ 0 What would you say is the best measure of central tendency for the weekly visitor distribution? How should the Department of Fish and Wildlife report it? Justify your response. I added all numbers up to equal 2,772 and divide by 40 which equals 69.3 which is the mean. The best measure of tendency is the mean because it is an average of all the numbers. It should be reported as the average amount of of visitors ages 65-81+ is 69.3. 3. A real estate agent has been given 10 houses to list. The house prices in dollars are: 110,000 95,000 87,000 92,000 99,000 250,000 265,000 210,000 275,000 240,000 A potential buyer calls and asks about house prices in the area. What measure of central tendency would best describe this data set? How should the real estate agent report it? Justify your response. 87,000, 92,000, 95,000,99,000, 110,000, 210,000, 240,000, 250,000, 265,000, 275,000 median=$160,000 The median best describes the central measure of tendency. If a single home that is high priced pulled up the average price well above a typical home on the market. This is why median provides a more accurate measure of the typical value of a home. She should report the median price in dollars. $160,000
4. Matthew collected the following data: Table 4.1 Effect of Diazinon Exposure on Tadpole Length Tadpole length (cm) Control 0.25 ppb 0.5 ppb 1.0 ppb Tadpole 1 4.4 4.4 3.1 3.5 Tadpole 2 4.6 3.5 3.4 3.1 Tadpole 3 4.2 3.8 3.7 2.8 Tadpole 4 3.8 4.2 4.0 3.3 Tadpole 5 4.3 4.1 3.3 2.6 Mean (cm) 4.26 4.0 3.5 3.06 Std Dev (cm) 0.3 0.4 0.4 0.4 t-test probability (p-value) NA 0.122 0.003 0.0002 Based on the data presented in the table above, are any of the treatment groups significantly different from the control group? If so, which groups are different and how do you know they are different? I could do a test to test for significance by using the significance value of 0.05 which is the standard significance level. If the p-value is greater than significance value than there is no difference in the treatment group. If the P-value is below the significance value of .05 than there is a difference in the treatment group. There is a significant difference in the 0.5 ppb and the 1.0 ppb and I know there different because their p-value is less than the level of significance . 05. 5. Natalie conducted a study on the effects of triclosan on three algae populations. She collected an algae sample from three different ponds, counted the number of algae cells present in the sample, and then exposed them to a 5 ppm aqueous solution of triclosan and re- counted the number of algae cells in each sample. Using a paired t-test, she obtained a p-value of 0.083 when comparing the original algae population sizes to the population sizes after exposure to triclosan. Natalie concluded that triclosan is toxic to algae because her data for the three samples showed a trend toward a reduction in population size. What could Natalie do to show a statistically significant effect of triclosan on algae population size? Justify your response. She would have to collect more data from the ponds or add more ponds. The p-value of .083 tells me that there is not a statistically significant effect of triclosan on algae populations. There would have to be a p-value of 0.05 or less to show a significant effect of triclosan on algae populations..
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