Copepods are tiny crustaceans that are an essential link in the estuarine food web. Marine scientists designed an experiment to determine whether dietary lipid Diatoms Bacteria Macroalgae429 292289(fat) content is important in the population growth of a copepod. Independent random samples of copepods were placed in containers containing lipid-richdiatoms, bacteria, or leafy macroalgae. There were 12 containers total with four replicates per diet. Five gravid (egg-bearing) females were placed in eachcontainer. After 14 days, the number of copepods in each container were as given to the right. At the 5% significance level, do the data provide sufficientevidence to conclude that a difference exists in mean number of copepods among the three different diets?465303311447295495329Click here to view page 1 of the F-distribution. Click here to view page 2 of the F-distribution. Click here to view page 3 of the F-distribution. Click here to view page 4 of the F-distribution.First, let μ₁, μ2, and μ3 be the population means times for diatoms, bacteria, and macroalgae, respectively. What are the correct hypotheses for a one-way ANOVA test?Ho:H₂:305278←

Given data: Diatoms Bacteria Macroalgae 429 292 289 465 303 311 447 295 305 495 329 278…

Answered: 429 292 Copepods are tiny crustaceans…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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An ornithologist thinks that injections of follicle-stimulating hormone (FSH) increase the singing rate of his captive male cotingas (birds). To test this hypothesis, he randomly selects 20 singing cotingas and divides them into two groups of 10 birds each. The first group receives injections of FSH and the second gets injections of saline solution, as a control for the trauma of receiving an injection. He then records the singing rate (in songs per hour) for both groups. The results are given in the following table. Note that two of the FSH birds escaped during injection and were not replaced. a.) What is the alternative hypothesis? Use a directional alternative hypothesis. b.) What is the null hypothesis? c.) Using the Mann-Whitney U test and alpha=0.05tail what is your conclusion?
It has been shown that the fertilizer magnesium ammonium phosphate, Mg, NH4PO4, is an effective supplier of the nutrients necessary for plant growth. A study was conducted at George Mason University to determine a possible optimum level of fertilization, based on the enhanced vertical growth response of the chrysanthemums. Forty chrysanthemum seedlings were divided into four groups, each containing 10 plants. Each was planted in a similar pot containing a uniform growth medium. To each group of plants an increasing concentration of Mg,NH4PO4, measured in grams per bushel, was added: 50 g/bu, 100 g/bu, 200 g/bu, and 400g/bu. The sample means for each group was 15.34 cm, 17.16 cm, 18.3 cm, and 20.1 cm, respectively. Here SST = 758.035. (a) Construct the ANOVA table. (b) Use the Bonferroni correction to construct all g pairwise confidence intervals with t a df 2.792. What is a?
Dr. Pagels is a mammalogist who studies meadow and common voles. He frequently traps the moles and has noticed what appears to be a preference for a peanut butter-oatmeal mixture by the meadow voles vs apple slices are usually used in traps, where the common voles seem to prefer the apple slices. So he conducted a study where he used a peanut butter-oatmeal mixture in half the traps and the normal apple slices in his remaining traps to see if there was a food preference between the two different voles. To test for independence, we need to calculate the Chi-square statistic. These are the data that Dr. Pagels collected ( the attachment): You will want to use Excel (or Google sheets) to do the calculations for you, so that your answers are exact (they need to be exact to get credit. Here is a Google sheet to get you started. When transferring your answers, make sure you carry them out to AT LEAST SIX SIGNIFICANT FIGURES unless otherwise stated. Round your final answer to 4 decimal…
A research focused on the number of illnesses involving the organism Escherichia coli (01257:H7), which causes a breakdown of red blood cells and intestinal hemorrhage in its victims. According to the Center for Disease Control of the Department of Health, an estimated 73,000 cases of E. coli infection and 61 deaths occur in Region Y. A 2028 outbreak traced to wild pigs, who spread the bacteria into a cabbage field in that , sickened 204 people in 26 municipalities. Outbreaks have occurred at a rate of 2.5 per 100,000. Let us suppose that this rate has not changed.a. What is the probability that at most five cases of E.coli per 100,000 are reported in California this year?b. What is the probability that more than five cases of E. coli are reported in California this year?c. Approximately 95% of occurrences of E. coli involve at most how many cases?
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A travel association says the daily lodging costs for a family in the United States is $132. You work for a tourist publication and want to test this claim. You randomly selected 10 U.S. families and find out how much each spent on lodging for one overnight trip. At alpha=0.02, can you reject the travel association's claim? (Adapted from the American Automobile Association) 164 137 142 155 119 104 74 204 148 181 Assume the population is normally distributed.
A geneticist working on peas has a single plant monohybrid Y/y (yellow) plant and, from a self of this plant,wants to produce a plant of genotype y/y to use as a tester. How many progeny plants need to be grown to be 95percent sure of obtaining at least one in the sample?
An agricultural scientist tests six types of fertilizer, labeled A, B, C, D, E, and F, to determine whether any of them produces an increase in the yield of lima beans over that obtained with the current fertilizer. For fertilizer C, the increase in yield is statistically significant at the 0.05 level. For the other five, the increase is not statistically significant. The scientist concludes that the yield obtained with fertilizer C is greater than that of the current fertilizer. Explain why this conclusion is not justified.
A study conducted by researchers was designed "to determine if application of duct tape is as effective as cryotherapy in the treatment of common warts." The researchers randomly divided 6969 patients into two groups. The 3535 patients in group 1 had their warts treated by applying duct tape. The 3434 patients in group 2 had their warts treated by cryotherapy. Once the treatments were complete, it was determined that 6969% of the patients in group 1 and 8585% of the patients in group 2 had complete resolution of their warts. The researchers concluded that cryotherapycryotherapy is significantly more effective in treating warts than duct tapeduct tape. Complete parts (a) through (d) below.(a) What is the research objective? A.To determine if duct tape is as effective as cryotherapy in the treatment of warts B.To determine if cryotherapy is as effective as duct tape in the treatment of warts C.To determine if cryotherapy is effective in the treatment of warts D.To determine if…
A study was conducted to demonstrate that soybeans inoculated with nitrogen-fixing bacteria yield more and grow adequately without expensive environmentally deleterious synthesized fertilizers. The trial was conducted under controlled conditions with uniform amounts of soil. The initial hypothesis was that inoculated plants would outperform their uninoculated counterparts. This assumption is based on the facts that plants need nitrogen to manufacture vital proteins and amino acids and that nitrogen-fixing bacteria would make more of this substance available to plants, increasing their size and yield. There were 8 inoculated plants (I) and 8 uninoculated plants (U). The plant yield as measured by pod weight for each plant is given in Table 2.20 Here are the questions I would like explanations for. Please use STATA to compute answers!!!: 2.35. The objective is to compute appropriate descriptive statistics for I and Data set available U plants. 2.36. Use graphic methods to compare the…
A study was conducted to demonstrate that soybeans inoculated with nitrogen-fixing bacteria yield more and grow adequately without expensive environmentally deleterious synthesized fertilizers. The trial was conducted under controlled conditions with uniform amounts of soil. The initial hypothesis was that inoculated plants would outperform their uninoculated counterparts. This assumption is based on the facts that plants need nitrogen to manufacture vital proteins and amino acids and that nitrogen-fixing bacteria would make more of this substance available to plants, increasing their size and yield. There were 8 inoculated plants (I) and 8 uninoculated plants (U). The plant yield as measured by pod weight for each plant is given in Table 2.20. Compute appropriate descriptive statistics for I and Data set available U plants.

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