![Elementary Statistics 2nd Edition](https://www.bartleby.com/isbn_cover_images/9781259724275/9781259724275_smallCoverImage.jpg)
Concept explainers
Exercises 21—24 refer to the population of animals in the following table. The population is divided into four groups: mammals, birds, reptiles, and fish.
Another sample: Draw a simple random sample of two groups of animals from the four groups, and construct a sample of 20 animals by including all the animals in the sampled groups. What kind of sample is this?
![Check Mark](/static/check-mark.png)
Tofind:The simple random sample of two groups of animal and the type of sample.
Answer to Problem 23E
The simple random sample of two groups of animal is
Explanation of Solution
Given information:The population is divided into four groups: mammals, birds, reptiles, and fish. Refer to the population of animals in the following table.
Mammals | Birds | Reptiles | Fish |
Aardvark Buffalo Elephant Squirrel Rabbit Lion Zebra Pig Dog Horse | 11. Flamingo 12. Swan 13. Sparrow 14. Parrot 15. Pelican 16. Hawk 17. Owl 18. Chicken 19. Duck 20. Turkey | 21. Gecko 22. Iguana 23. Chameleon 24. Rattlesnake 25. Boa constrictor 26. Python 27. Turtle 28. Tortoise 29. Alligator 30. Crocodile | 31. Catfish 32. Tuna 33. Cod 34. Salmon 35. Goldfish 36. Shark 37. Trout 38. Perch 39. Guppy 40. Minnow |
Concept Involved:
If a sample is taken from a population and each item is equally like to make the sample then the sample is called simple random sample.
Calculation:
Since, the items are drawn from the population in groups, or clusters.
Thus, the type of sample is cluster sample.
Consider a simple random sample of two groups of animals from the four groups, and construct a sample of 20 animals by including all the animals in the sampled groups.
Group | Samples | |
Mammals | 1 | Aardvark, Buffalo, Elephant, Squirrel, Rabbit, Lion, Zebra, Pig, Dog, Horse |
Birds | 2 | Flamingo, Swan, Sparrow, Parrot, Pelican, Hawk, Owl, Chicken, Duck, Turkey |
Reptiles | 3 | Gecko, Iguana, Chameleon, Rattlesnake, Boa constrictor, Python, Turtle, Tortoise, Alligator, Crocodile |
Fish | 4 | Catfish, Tuna, Cod, Salmon, Goldfish, Shark, Trout, Perch, Guppy, Minnow |
The step-by-step procedure is shown below.
Step 1: Enter any nonzero number on the HOME screen as the seed Step 2: Press Step 3: Press Step 4: Press Then enter Step 5: Press |
Therefore, the simple random sample of two groups of animal is
Want to see more full solutions like this?
Chapter 1 Solutions
Elementary Statistics 2nd Edition
Additional Math Textbook Solutions
Pathways To Math Literacy (looseleaf)
Precalculus: Mathematics for Calculus (Standalone Book)
Elementary and Intermediate Algebra: Concepts and Applications (7th Edition)
Graphical Approach To College Algebra
Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)
Probability And Statistical Inference (10th Edition)
- Stem1: 1,4 Stem 2: 2,4,8 Stem3: 2,4 Stem4: 0,1,6,8 Stem5: 0,1,2,3,9 Stem 6: 2,2 What’s the Min,Q1, Med,Q3,Max?arrow_forwardAre the t-statistics here greater than 1.96? What do you conclude? colgPA= 1.39+0.412 hsGPA (.33) (0.094) Find the P valuearrow_forwardA poll before the elections showed that in a given sample 79% of people vote for candidate C. How many people should be interviewed so that the pollsters can be 99% sure that from 75% to 83% of the population will vote for candidate C? Round your answer to the whole number.arrow_forward
- Suppose a random sample of 459 married couples found that 307 had two or more personality preferences in common. In another random sample of 471 married couples, it was found that only 31 had no preferences in common. Let p1 be the population proportion of all married couples who have two or more personality preferences in common. Let p2 be the population proportion of all married couples who have no personality preferences in common. Find a95% confidence interval for . Round your answer to three decimal places.arrow_forwardA history teacher interviewed a random sample of 80 students about their preferences in learning activities outside of school and whether they are considering watching a historical movie at the cinema. 69 answered that they would like to go to the cinema. Let p represent the proportion of students who want to watch a historical movie. Determine the maximal margin of error. Use α = 0.05. Round your answer to three decimal places. arrow_forwardA random sample of medical files is used to estimate the proportion p of all people who have blood type B. If you have no preliminary estimate for p, how many medical files should you include in a random sample in order to be 99% sure that the point estimate will be within a distance of 0.07 from p? Round your answer to the next higher whole number.arrow_forward
- A clinical study is designed to assess the average length of hospital stay of patients who underwent surgery. A preliminary study of a random sample of 70 surgery patients’ records showed that the standard deviation of the lengths of stay of all surgery patients is 7.5 days. How large should a sample to estimate the desired mean to within 1 day at 95% confidence? Round your answer to the whole number.arrow_forwardA clinical study is designed to assess the average length of hospital stay of patients who underwent surgery. A preliminary study of a random sample of 70 surgery patients’ records showed that the standard deviation of the lengths of stay of all surgery patients is 7.5 days. How large should a sample to estimate the desired mean to within 1 day at 95% confidence? Round your answer to the whole number.arrow_forwardIn the experiment a sample of subjects is drawn of people who have an elbow surgery. Each of the people included in the sample was interviewed about their health status and measurements were taken before and after surgery. Are the measurements before and after the operation independent or dependent samples?arrow_forward
- iid 1. The CLT provides an approximate sampling distribution for the arithmetic average Ỹ of a random sample Y₁, . . ., Yn f(y). The parameters of the approximate sampling distribution depend on the mean and variance of the underlying random variables (i.e., the population mean and variance). The approximation can be written to emphasize this, using the expec- tation and variance of one of the random variables in the sample instead of the parameters μ, 02: YNEY, · (1 (EY,, varyi n For the following population distributions f, write the approximate distribution of the sample mean. (a) Exponential with rate ẞ: f(y) = ß exp{−ßy} 1 (b) Chi-square with degrees of freedom: f(y) = ( 4 ) 2 y = exp { — ½/ } г( (c) Poisson with rate λ: P(Y = y) = exp(-\} > y! y²arrow_forward2. Let Y₁,……., Y be a random sample with common mean μ and common variance σ². Use the CLT to write an expression approximating the CDF P(Ỹ ≤ x) in terms of µ, σ² and n, and the standard normal CDF Fz(·).arrow_forwardmatharrow_forward
- Holt Mcdougal Larson Pre-algebra: Student Edition...AlgebraISBN:9780547587776Author:HOLT MCDOUGALPublisher:HOLT MCDOUGALGlencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw HillBig Ideas Math A Bridge To Success Algebra 1: Stu...AlgebraISBN:9781680331141Author:HOUGHTON MIFFLIN HARCOURTPublisher:Houghton Mifflin Harcourt
- College Algebra (MindTap Course List)AlgebraISBN:9781305652231Author:R. David Gustafson, Jeff HughesPublisher:Cengage Learning
![Text book image](https://www.bartleby.com/isbn_cover_images/9780547587776/9780547587776_smallCoverImage.jpg)
![Text book image](https://www.bartleby.com/isbn_cover_images/9780079039897/9780079039897_smallCoverImage.jpg)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781680331141/9781680331141_smallCoverImage.jpg)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305652231/9781305652231_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781337282291/9781337282291_smallCoverImage.gif)