1. You would like to estimate bacterial levels (specifically Escherichia coli) for a large shallowpond. You collect 1-L water samples from 5 random locations in the pond. You measure E. colilevels using 10-mL subsamples (three per 1-L water sample).(a) This sampling design would be an example of what type of sampling strategy?(b) Table 1 shows the data that you collected. Using equations appropriate to your samplingstrategy, calculate the mean E. coli concentration for the pond and estimate the standard error.The estimated volume of the pond is 10,000 m³ or 1×10' L, such that the total number ofpossible sample units is N= pond volume / sample volume = 1×107 L /1 L = 1×107. (Note,that fi = n/N is very small!)Table 1. Measured E. coli levels (in CFU/100 mL) in sample subsamples.Also indicated are the sample means (y;) and sample sum-of-squares(SSy; i.e. E(yy – y.)³,.Recalculate the(c)mean, treating eachsubsample as a sample(i.e. Eyi /15, where2yi is the sum of allmeasured values). Howdoes this mean compareSubsamplesSy2,400Sample123Yi120808060.05027050123.332,300312017096.715,30047023.33,2705110403060.03,800to the mean youcalculated in (b)?(d) Assume one of the subsamples was lost (so unequal number subsamples across the samples)and you recalculated the mean as in (b) and (c). How might you expect the two meancalculations to compare? Which version should be a better estimate of the true populationmean? Why? [hint: try removing subsample 1 of sample 4]

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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 agricultural field trial compares the yield of two varieties of corn. The researchers divide in half each of 20 fields of land in different locations and plant each corn variety in one half of each plot. After harvest, the yields are compared in bushels per acre at each location. The 20 differences (Variety A - Variety B) givex¯=3.2x¯=3.2 and s=4.63s=4.63. Does this sample provide evidence that Variety A had a higher yield than Variety B? (a) State the null and alternative hypotheses: (Type "mu" for the symbol μμ , e.g. mu >> 1 for the mean is greater than 1, mu << 1 for the mean is less than 1, mu not = 1 for the mean is not equal to 1) H0H0 : HaHa : (b) Find the test statistic, t = (c) Answer the question: Does this sample provide evidence that Variety A had a higher yield than Variety B? (Use a 5% level of significance) (Type: Yes or No)
Using the same question from #3 A survey is done to find out the favorite movies of 500 children. The children are categorized into male and female, and a sample of 50 children is chosen. Answer the following question: If the number of female children is 300, what is the sample size of each stratum? a male sample: 40 and female sample: 10 b male sample; 25 and female sample: 25 c male sample: 30 and female sample: 20 d male sample: 20 and female sample: 30
Bighorn sheep are beautiful wild animals found throughout the western United States. Let x be the age of a bighorn sheep (in years), and let y be the mortality rate (percent that die) for this age group. For example, x = 1, y = 14 means that 14% of the bighorn sheep between 1 and 2 years old died. A random sample of Arizona bighorn sheep gave the following information: x 1 2 3 4 5 y 15.8 17.3 14.4 19.6 20.0 Σx = 15; Σy = 87.1; Σx2 = 55; Σy2 = 1,540.45; Σxy = 272 (a) Find x, y, b, and the equation of the least-squares line. (Round your answers for x and y to two decimal places. Round your least-squares estimates to three decimal places.) x = y = b = ŷ = + x (b) Draw a scatter diagram for the data. Plot the least-squares line on your scatter diagram. (c) Find the sample correlation coefficient r and the coefficient of determination r2. (Round your answers to three decimal places.) r = r2 = What percentage of variation in y is…
A psychologist would like to examine the effects of a new drug on the activity level of animals. Three samples of rats are selected with n=5 in each sample. One group gets no drug, one group gets s small dose, and the third group gets a large dose. The psychologist records the activity level for each animal. The data from this experiment are presented below. No drug Small dose Large dose 5, 3 , 1 , 1 , 5 5, 5, 9, 6, 5 10, 12, 9, 6, 8 =3, SS=16 =6, SS=12 =9, SS=20 Do these data indicate any significant differences among the three groups? Test with α=0.05. Compute the effect size and interpret the result.
The control department of a light bulb manufacturer randomly picks 4400 light bulbs from the production lot every week. The records show that, when there is no malfunction, the defect rate in the manufacturing process (due to imperfections in the material used) is 1% . When 1.5% or more of the light bulbs in the sample of 4400 are defective, the control unit calls repair technicians for service. Answer the following. (If necessary, consult a list of formulas.) (a)Find the mean of p , where p is the proportion of defective light bulbs in a sample of 4400 when there is no malfunction. (b)Find the standard deviation of p . (c)Compute an approximation for P≥p0.015 , which is the probability that the service technicians will be called even though the system is functioning properly. Round your answer to four decimal places.
Can a person train to become better at holding their breath? An experiment was designed to find out. Twelve volunteers were randomly assigned to 1 of 2 groups. The 6 volunteers assigned to group 1 were given breath-holding exercises to perform for 2 weeks. The other group was not given any information about the experiment. At the end of the 2 weeks, all 12 volunteers were individually tested to determine how long they could hold their breath. Here are the data (in seconds). Group 1: 90, 88, 70, 110, 75, 105 Group 2: 40, 48, 35, 50, 55, 62 The researcher would like to determine if these data provide convincing evidence that the true mean amount of time volunteers who were given training held their breath is greater than volunteers without training. Let ₁ = the true mean amount of time that volunteers who were given training held their breath and ₂ = the true mean amount of time that volunteers without training held their breath. What are the appropriate hypotheses? Ho: U1 - U2 = 0, Ha:…
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Bighorn sheep are beautiful wild animals found throughout the western United States. Let x be the age of a bighorn sheep (in years), and let y be the mortality rate (percent that die) for this age group. For example, x = 1, y = 14 means that 14% of the bighorn sheep between 1 and 2 years old died. A random sample of Arizona bighorn sheep gave the following information: x 1 2 3 4 5 y 15.8 17.3 14.4 19.6 20.0 Σx = 15; Σy = 87.1; Σx2 = 55; Σy2 = 1,540.45; Σxy = 272 (a) Find x, y, b, and the equation of the least-squares line. (Round your answers for x and y to two decimal places. Round your least-squares estimates to three decimal places.) x = y = b = ŷ = + x
You may need to use the appropriate technology to answer this question. An automobile dealer conducted a test to determine if the time in minutes needed to complete a minor engine tune-up depends on whether a computerized engine analyzer or an electronic analyzer is used. Because tune-up time varies among compact, intermediate, and full-sized cars, the three types of cars were used as blocks in the experiment. The data obtained follow. Analyzer Computerized Electronic Compact 50 41 Car Intermediate 54 44 Full-sized 64 47 Use a = 0.05 to test for any significant differences. State the null and alternative hypotheses. O Ho: MCompact = "Intermediate = HFull-sized H: "Compact * "Intermediate * "Full-sized O Ho: "Compact * "Intermediate * HFull-sized H: "Compact "Intermediate = "Full-sized O Ho: Computerized = HElectronic H: "Computerized * HElectronic O Ho: "Computerized = HElectronic = "Compact = Intermediate = Full-sized H: Not all the population means are equal. O Ho: HComputerized *…
Several methods of estimating the number of seeds in soil samples have been developed by ecologists. An article gave the accompanying data on the number of seeds detected by the direct method and by the stratified method for 27 soil specimens. Specimen 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Direct 21 33 0 60 20 61 40 8 13 93 1 68 77 21 Stratified 8 37 8 57 53 61 28 8 8 100 0 57 68 53 Specimen 15 Find the test statistic. (Round your answer to two decimal places.) t = 16 17 18 19 20 21 22 23 24 25 26 27 Direct Find the df. (Round your answer down to the nearest whole number.) df = 33 0 37 17 93 1 40 21 0 8 13 Use technology to find the P-value. (Round your answer to four decimal places.) P-value = 17 40 Stratified 28 0 37 13 93 13 48 21 0 13 40 Do the data provide sufficient evidence to conclude that the mean number of seeds detected differs for the two methods? Test the relevant hypotheses using a = 0.05. (Use direct stratified.) 13 77
The authors of a paper compared two different methods for measuring body fat percentage. One method uses ultrasound, and the other method uses X-ray technology. Body fat percentages using each of these methods for 16 athletes (a subset of the data given in a graph that appeared in the paper) are given in the accompanying table. You can assume that the 16 athletes who participated in this study are representative of the population of athletes. Athlete X-ray Ultrasound 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 5.00 8.00 9.25 12.00 17.25 29.50 5.50 6.00 8.00 13.50 9.25 11.00 12.00 14.00 17.00 18.00 4.25 4.75 9.00 11.75 17.00 27.50 6.50 6.75 8.75 14.50 9.50 12.00 12.25 15.50 18.00 18.25 Use these data to estimate the difference in mean body fat percentage measurement for the two methods. Use a confidence level of 95%. (Use μ = MX-ray-Multrasound. Round your answers to three decimal places.) × % Interpret the interval in context. O There is a 95% chance that the true mean body fat percentage…
Prices of diamonds are very sensitive to their weight (measured in carat). For example, the price of a 1 carat diamond tends to be much higher than the price of a 0.99 carats diamond. To study this phenomenon in more detail, we consider two random samples of diamonds, 0.99 carats and 1 carat, each sample of size 23. The average price of 0.99 carats diamonds in the sample is $4451, while the average price of 1 carat diamonds in the sample is $5681. From long-term studies, we know that both the prices of 0.99 carats diamonds and the prices of 1 carat diamonds are normally distributed with standard deviations oj = $1332 (0.99 carats) and o2 = $1613 (1 carat), respectively. (a) Carry out a hypothesis test at significance level 0.01 to evaluate whether the true average price µi of 0.99 carats diamonds is lower than the true average price µ2 of 1 carat diamonds. Specify Ho and Ha, the test statistic, and calculate the p-value or use p-value considerations to conclude. (b) Compute the type II…

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