5. Oxygen demand Is a term blologists use to describe the axygen needed by fish and other aquatic organisms for survival. The Environmental Protection Agency conductec study of a wetland area in Marin County, Californla. In this wetland environment, the m oxygen demand was u -9.9 mg/L, and a 1.7 mg/L. (Reference: EPA Report 832-R 005). Let x be a random variable that represents oxygen demand in this wetland environment. Assume x has a probability distribution that is approximately normal. a. An oxygen demand below 8 mg/L Indicates that some organisms in the wetlanc environment may be dying. What is the probability that the oxygen demand will fa

5. Oxygen demand Is a term blologists use to describe the axygen needed by fish and other aquatic organisms for survival. The Environmental Protection Agency conductec study of a wetland area in Marin County, Californla. In this wetland environment, the m oxygen demand was u -9.9 mg/L, and a 1.7 mg/L. (Reference: EPA Report 832-R 005). Let x be a random variable that represents oxygen demand in this wetland environment. Assume x has a probability distribution that is approximately normal. a. An oxygen demand below 8 mg/L Indicates that some organisms in the wetlanc environment may be dying. What is the probability that the oxygen demand will fa

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Description: 5. Oxygen demand is a term blologists use to describe the axygen needed by fish andother aquatic organisms for survval. The Environmental Protection Agency conducted astudy of a wetland area in Marin County, Californla. In this wetland environment,

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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FIV (feline immunodeficiency virus) attacks a cat's immune system causing FAIDS (feline acquired immunodeficiency syndrome). It is estimated that between 2 and 25 percent of the global domestic cat population is infected with the virus. FIV is primarily transmitted from cat to cat through biting. A one-step immunochromatographic (IC) test has been developed for the detection of FIV antibodies. It is estimated that 94.7% of cats with FIV test positive and 96.1% of cats without FIV test negative using this test. Assume that 5% of the local cat population is actually infected with FIV. Use this information to answer the following: (a) For a population of 100 000 cats: (i) how many are estimated to be infected with FIV? (ii) how many of the cats in (a)(i) will test positive? (b) Use your answers from (a) to help construct a 2 × 2 table of counts displaying the results for this sample. Complete the table. Estimate the proportion of local cats that would test negative for FIV but actually be…
The data in the attached image represents blood pressure based on a new trial medication used in an experiment involving a total of 50 male and female subjects. There were two levels of treatment: 0 = placebo and 1 = treatment group. Family History Blood Pressure = Fam-Hist-BP; from the paternal side = PS; from the maternal side = MS; from both sides = BS; none on both sides or not known = N.; Blood pressure before the experiment = BP-Before-Exp; Blood pressure after the experiment = BP-After-Exp. Task: Generate at least two different cross tabulations.
A research study investigated the composition of wines from the Rhine and Moselle wine regionsof Germany. One component measured was the concentration of i-Butyl alcohol. The results ofthe study are listed in the following table:Region n i-Butyl alcohol (mg/100ml)Rhine 9 ̄x =5.32,s =2.24Moselle 14 ̄x =3.62,s =0.97(a) Assume that the underlying population variances from the two regions are identical. At the0.05 level of significance, test the null hypothesis that the mean concentration of i-Butylalcohol is the same for wines from the Rhine and Moselle regions.Construct a 95% confidence interval(b) Now assume that the underlying population variances from the two regions differ. At the 0.05level of significance, test the null hypothesis that the mean concentration of i-Butyl alcohol isthe same for wines from the Rhine and Moselle regions, construct a 95% confidence interval(c) If you were to provide a report to the investigators which of the above results would youinclude and why?
Specific Motors of Detroit has developed a new automobile known as the M car. 24 M cars and 28 J cars (from Japan) were road tested to compare miles-per-gallon (mpg) performance. sample 1 (M cars) : n1 = 24, X ̄1 = 29.8, s1 = 2.56 sample 2 (J cars) : n2 = 28, X ̄2 = 27.3, s2 = 1.81 (b) Can we conclude, using a .05 level of significance, that the miles-per-gallon (mpg) performance of M cars is greater than the miles-per-gallon performance of J cars? (d) Find the critical values. (e) Rejection region and conclusion. (f) Interpret the results.
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A survey is conducted to determine if there is a difference in the proportion of students, parents, and The conditions for inference are met. The chi-square test statistic is 0.19 and df = 2. What is the appropriate decision using a = 0.05? teachers who volunteer at least once a month. To investigate, a random sample of 45 students, 25 parents, and 12 teachers was selected from a large high school. The data are displayed in the table. Find the chi-square table here. Because the P-value is less than a = 0.05, the null hypothesis should be rejected. Student Parent Teacher Because the P-value is less than a = 0.05, the null hypothesis should not be rejected. Yes 24 12 Volunteer? No 21 13 Because the P-value is greater than a = 0.05, the null hypothesis should be rejected. The researcher would like to test these hypotheses: Because the P-value is greater than a = 0.05, the null hypothesis should not be rejected. Họ: There is no difference in the distribution of responses among these three…
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Patients with rheumatoid arthritis are at a greater risk of developing osteoporosis. The reasons are not well understood due to the difficulty in qualitatively assessing bone metabolism and mineral content. A researcher measures human calcitonin (HCT) levels in men with rheumatoidarthritis. The researcher randomly selects 43 men with rheumatoid arthritis and measured the mean HCT level of 33.8 pg/mL. Assume that HCT for individuals with rheumatoid arthritis follows a normal distribution with standard deviation σ = 23.6 pg/mL. 3. Create a 95% confidence interval for the population mean HCT.
The data in table is from a randomized block design represent blood pressure measurements from an experiment involving 4 age groups, each with 3 persons. The 3 persons within each age group were randomly assigned to drugs, A, B and C with one person per drug. This was done to keep the drug assignments balanced within age groups. Comparing the three drugs in manner that provides balance for or control for possible age effects. Consider the data given in table below and compute; a. Mean Square Due to Treatments (MSTR) b. Mean Square Due to Blocks (MSBL) Mean Square Due to Error (MSE) C. d. F-test Treatments Age Groups (Blocks) A B 1 70 72 2 76 84 3 82 86 4 90 92 Total 318 334 C Total Blocks 80 222 82 242 84 252 88 270 334 986