**10.12** An entomologist is studying a new tick species that may be the carrier of the pathogen associated with Lyme disease. She designs a study to estimate the prevalence of the pathogen in the tick. She examines 100 ticks randomly selected in the study region during a period of the year when ticks have been known to be infected with the pathogen in other regions of the country. The examination of the 100 ticks finds none of the ticks are infected with the pathogen. a. Provide the entomologist with an estimate of the proportion of ticks of this species that are carrying the pathogen.b. Construct a 95% confidence interval for the proportion of ticks of this species that are carrying the pathogen.

The total number of randomly selected sample of ticks, n=100The number of ticks that are infected…

Answered: 10.12 An entomologist is studying a new…

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 official of a large car manufacturing company wanted to study the relationship between an employee's age and the number of absences. For this, a random sample of 10 randomly selected employees was collected and the following data was obtained. The material: age 40 28 34 27 21 38 19 55 31 35 number of absences 1 6 6 9 12 4 13 2 5 3 (a) Predict the number of absenteeism for a 42-year-old employee.(b) Determine the 95% prediction interval for the number of absences for the employee in (a).
Amorphophallus johnsonii is a plant called the "corpse flower" that grows in West Africa. Its flowers produce a pungent smell described as "a powerful aroma of rotting fish and faeces". It does so to attract its pollinator, the carrion beetle (Phaeochrous amplus) which live in rotting meat. Beath (1996) observed corpse flowers (hopefully wearing masks) and counted the number of beetles that arrived at each flower over the course of one night. The data is presented below: 51, 45, 61, 76, 11, 117, 7, 132, 52, 149 the mean of this sample is 70.1 beetles the standard deviation of this sample is 48.5 beetles Use the standard deviation of the sample to approximate the standard error of the mean. 1) What is the approximate SE that you calculated? Use the SE approximation method for normal sampling distributions to calculate Confidence Intervals 2) What are the 95% Confidence Intervals for this sample mean? 3) What are the 99% Confidence Intervals for this sample mean? 4) If an expanded…
An experiment was conducted to test the effects of five different diets (A-E) in turkeys. Six turkeys were randomly assigned to each of the five different diet groups (Diet) and were fed for a fixed period of time. The weight gained in pounds was recorded for each turkey and the data was entered into R for analysis. R produced the following output. Between Within Total a. له له له م په b. C. d. e. Sum-of-squares (SS) Degrees of freedom Mean square (MS) ?? (b) 103.04 ???.?? (a) 110.9249 25 29 ????.??? (c) ????.??? (d) Find the sum-of-squares for within-group variability. Find the degrees of freedom for between-group variability. Find the mean-square of the between-group variability. F-statistic ??.????? p-value ????.??? (e) Find the mean-square of the within-group variability. Find the p-value for the ANOVA F-test. What is the conclusion? Are the means of the five diets significantly different?
During the 2013 season, Major League Baseball took steps to speed up the play of baseball games in order to maintain fan interest. The average duration of a game from a sample of 60 games played during the summer of 2012 was 2 hours and 52 minutes. The average duration of a game from a sample of 50 games played during the summer of 2013 was 2 hours and 46 minutes. For the purposes of this question, let 2013 games be population 1. 1a. A research hypothesis was that the steps taken during the 2013 season would reduce the population mean duration of baseball games. Formulate null and alternative hypotheses. 1b. What is the point estimate of the reduction in the mean duration of games during the 2003 season?
A study of fox rabies in a country gave the following information about different regions and the occurrence of rabies in each region. A random sample of n1 = 16 locations in region I gave the following information about the number of cases of fox rabies near that location. x1: Region I Data 2 8 8 8 6 8 8 1 3 3 3 2 5 1 4 6 A second random sample of n2 = 15 locations in region II gave the following information about the number of cases of fox rabies near that location. x2: Region II Data 1 1 5 1 4 8 5 4 4 4 2 2 5 6 9 (i) Use a calculator with sample mean and sample standard deviation keys to calculate x1 and s1 in region I, and x2 and s2 in region II. (Round your answers to two decimal places.) x1 = s1 = x2 = s2 = (ii) Does this information indicate that there is a difference (either way) in the mean number of cases of fox rabies between the two regions? Use a 5% level of significance. (Assume the distribution of rabies cases in both regions is…
Do lizards play a role in spreading plant seeds? Some research carried out in a country would suggest so. The researchers collected 400 seeds of this particular type of fig, 100 of which were from each treatment: lizard dung, bird dung, rock hyrax dung, and uneaten figs. They planted these seeds in batches of 5, and for each group of 5 they recorded how many of the seeds germinated. This resulted in 20 observations for each treatment. The treatment means and standard deviations are given in the accompanying table. Treatment n Uneaten figs 20 2.80 0.30 Lizard dung 20 2.75 0.37 Bird dung 20 2.10 0.35 Hyrax dung 20 1.85 0.26 n USE SALT (a) Construct the appropriate ANOVA table, and test the hypothesis that there is no difference between the means for the number of seeds germinating for the four treatments. (Use a = 0.05. Round your mean squares to three decimal places and F statistic to two decimal places.) Source of Sum of Mean df Variation Squares Square Treatments 3 26.25 8.750 83.93…
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A study was conducted to determine whether big-city and small-town dwellers differed in their helpfulness to strangers. In this study, the investigators rang the doorbells of strangers living in a large City or small towns in the vicinity. They explained they had misplaced the address of a friend living in the neighbourhood and asked to use the phone. The following data show the number of individuals who admitted or did not admit the strangers (the investigators) into their homes: Helpfulness to strangers Admitted strangers into their home Didnot admit strangers into their home Big city dwellers 60 90 Small town dwellers 70 30 State the dependent and independent variable Is this a directional or non directional
A study of fox rabies in southern Germany gave the following information about different regions and the occurrence of rabies in each region. A random sample of n1 = 16 locations in region 1 gave the following information about the number of cases of fox rabies near that location. x1: Region I Data 3 7 8 7 5 8 8 1 3 3 3 2 5 1 4 6 A second random sample of n2 = 15 locations in region II gave the following information about the number of cases of fox rabies near that location. x2: Region II Data 4 2 4 1 3 8 5 4 4 4 2 2 5 6 9 Note: If a two-sample t-test is appropriate, for degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value by a small amount and therefore produce a slightly more "conservative" answer.Use a calculator with sample mean and sample standard deviation keys to find x1 and s1 in region I, and x2 and s2 in region II. (Round your answers to two…
Refer to Exercise 8.S.6. Analyze these data using a Wilcoxon signed-rank test.
Q.6.2 The financlal manager of a natlonal warehouse chaln has found from prevlous experlence that 45% of clients pay by credit card, 35% pay on account, whille the remainder pay using cash. He wishes to know if this is still the case. A random sample of 1000 sales receipts from the last month produced the following results: Payment Method Credit Card Account Cash Total Number of receipts 500 315 185 1000 Using a 1% significance level, test whether the payment method pattern has changed.

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