PRACTICE OF STATISTICS F/AP EXAM
PRACTICE OF STATISTICS F/AP EXAM
6th Edition
ISBN: 9781319113339
Author: Starnes
Publisher: MAC HIGHER
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Chapter 4.1, Problem 15E

(a)

To determine

To explain why it would not be practical for scientists to obtain an SRS in this setting.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given that:

Total pine trees =5000

Population size =5000

Sample pine trees =200

This implies sample size =200

Now, in case of 5000 pine trees which is spread around highway.

Now, we have to select 200 trees from that but practically we select more trees near by highway than trees which are from highway or vice versa. But practically we do not select 200 trees such that every trees has equal chance of selection. Thus, probability of selecting every treesis not same. Therefore, according to definition of SRS, it would not be practical for scientist to obtain an SRS in this setting.

(b)

To determine

To explain why is this sampling method biased.

(b)

Expert Solution
Check Mark

Explanation of Solution

A possible alternative would be to use first 200 pine trees along the highway as you enter the park.

In this case, we only select first 200 pine trees. So, it is possible that first 200 trees are not so much infected than remaining tree or it is also possible that the infection pine beetles is more near the highway than other trees.

So, if we only study about first 200 trees than there is bias for result and we do not get perfect result for study. Therefore, this idea is good.

(c)

To determine

To find out can scientist conclude that exactly 35% of all the pine trees on the west side of the park are infected and explain why or why not.

(c)

Expert Solution
Check Mark

Answer to Problem 15E

Yes.

Explanation of Solution

Systematic sampling: in this sampling starting from a random point on a sampling frame, every nth element in the frame is selected at equal interval.

In this study, total trees =5000

And sample pine trees =200

Therefore, interval for selection,

  5000200=25

It means we select every 25 th tree from starting at first three and end at 5000 th tree. So, we get 200 trees at interval of 25 trees from 5000 tree. In this sampling entire population cover. So, study of this sampling give accurate result for infection in trees and proportion of infection in trees.

Chapter 4 Solutions

PRACTICE OF STATISTICS F/AP EXAM

Ch. 4.1 - Prob. 11ECh. 4.1 - Prob. 12ECh. 4.1 - Prob. 13ECh. 4.1 - Prob. 14ECh. 4.1 - Prob. 15ECh. 4.1 - Prob. 16ECh. 4.1 - Prob. 17ECh. 4.1 - Prob. 18ECh. 4.1 - Prob. 19ECh. 4.1 - Prob. 20ECh. 4.1 - Prob. 21ECh. 4.1 - Prob. 22ECh. 4.1 - Prob. 23ECh. 4.1 - Prob. 24ECh. 4.1 - Prob. 25ECh. 4.1 - Prob. 26ECh. 4.1 - Prob. 27ECh. 4.1 - Prob. 28ECh. 4.1 - Prob. 29ECh. 4.1 - Prob. 30ECh. 4.1 - Prob. 31ECh. 4.1 - Prob. 32ECh. 4.1 - Prob. 33ECh. 4.1 - Prob. 34ECh. 4.1 - Prob. 35ECh. 4.1 - Prob. 36ECh. 4.1 - Prob. 37ECh. 4.1 - Prob. 38ECh. 4.1 - Prob. 39ECh. 4.1 - Prob. 40ECh. 4.1 - Prob. 41ECh. 4.1 - Prob. 42ECh. 4.2 - Prob. 43ECh. 4.2 - Prob. 44ECh. 4.2 - Prob. 45ECh. 4.2 - Prob. 46ECh. 4.2 - Prob. 47ECh. 4.2 - Prob. 48ECh. 4.2 - Prob. 49ECh. 4.2 - Prob. 50ECh. 4.2 - Prob. 51ECh. 4.2 - Prob. 52ECh. 4.2 - Prob. 53ECh. 4.2 - Prob. 54ECh. 4.2 - Prob. 55ECh. 4.2 - Prob. 56ECh. 4.2 - Prob. 57ECh. 4.2 - Prob. 58ECh. 4.2 - Prob. 59ECh. 4.2 - Prob. 60ECh. 4.2 - Prob. 61ECh. 4.2 - Prob. 62ECh. 4.2 - Prob. 63ECh. 4.2 - Prob. 64ECh. 4.2 - Prob. 65ECh. 4.2 - Prob. 66ECh. 4.2 - Prob. 67ECh. 4.2 - Prob. 68ECh. 4.2 - Prob. 69ECh. 4.2 - Prob. 70ECh. 4.2 - Prob. 71ECh. 4.2 - Prob. 72ECh. 4.2 - Prob. 73ECh. 4.2 - Prob. 74ECh. 4.2 - Prob. 75ECh. 4.2 - Prob. 76ECh. 4.2 - Prob. 77ECh. 4.2 - Prob. 78ECh. 4.2 - Prob. 79ECh. 4.2 - Prob. 80ECh. 4.2 - Prob. 81ECh. 4.2 - Prob. 82ECh. 4.2 - Prob. 83ECh. 4.2 - Prob. 84ECh. 4.2 - Prob. 85ECh. 4.2 - Prob. 86ECh. 4.2 - Prob. 87ECh. 4.2 - Prob. 88ECh. 4.2 - Prob. 89ECh. 4.2 - Prob. 90ECh. 4.2 - Prob. 91ECh. 4.2 - Prob. 92ECh. 4.3 - Prob. 93ECh. 4.3 - Prob. 94ECh. 4.3 - Prob. 95ECh. 4.3 - Prob. 96ECh. 4.3 - Prob. 97ECh. 4.3 - Prob. 98ECh. 4.3 - Prob. 99ECh. 4.3 - Prob. 100ECh. 4.3 - Prob. 101ECh. 4.3 - Prob. 102ECh. 4.3 - Prob. 103ECh. 4.3 - Prob. 104ECh. 4.3 - Prob. 105ECh. 4.3 - Prob. 106ECh. 4.3 - Prob. 107ECh. 4.3 - Prob. 108ECh. 4.3 - Prob. 109ECh. 4.3 - Prob. 110ECh. 4.3 - Prob. 111ECh. 4.3 - Prob. 112ECh. 4.3 - Prob. 113ECh. 4.3 - Prob. 114ECh. 4.3 - Prob. 115ECh. 4.3 - Prob. 116ECh. 4.3 - Prob. 117ECh. 4.3 - Prob. 118ECh. 4.3 - Prob. 119ECh. 4.3 - Prob. 120ECh. 4 - Prob. R4.1RECh. 4 - Prob. R4.2RECh. 4 - Prob. R4.3RECh. 4 - Prob. R4.4RECh. 4 - Prob. R4.5RECh. 4 - Prob. R4.6RECh. 4 - Prob. R4.7RECh. 4 - Prob. R4.8RECh. 4 - Prob. R4.9RECh. 4 - Prob. R4.10RECh. 4 - Prob. T4.1SPTCh. 4 - Prob. T4.2SPTCh. 4 - Prob. T4.3SPTCh. 4 - Prob. T4.4SPTCh. 4 - Prob. T4.5SPTCh. 4 - Prob. T4.6SPTCh. 4 - Prob. T4.7SPTCh. 4 - Prob. T4.8SPTCh. 4 - Prob. T4.9SPTCh. 4 - Prob. T4.10SPTCh. 4 - Prob. T4.11SPTCh. 4 - Prob. T4.12SPTCh. 4 - Prob. T4.13SPTCh. 4 - Prob. T4.14SPTCh. 4 - Prob. AP1.1CPTCh. 4 - Prob. AP1.2CPTCh. 4 - Prob. AP1.3CPTCh. 4 - Prob. AP1.4CPTCh. 4 - Prob. AP1.5CPTCh. 4 - Prob. AP1.6CPTCh. 4 - Prob. AP1.7CPTCh. 4 - Prob. AP1.8CPTCh. 4 - Prob. AP1.9CPTCh. 4 - Prob. AP1.10CPTCh. 4 - Prob. AP1.11CPTCh. 4 - Prob. AP1.12CPTCh. 4 - Prob. AP1.13CPTCh. 4 - Prob. AP1.14CPTCh. 4 - Prob. AP1.15CPTCh. 4 - Prob. AP1.16CPTCh. 4 - Prob. AP1.17CPTCh. 4 - Prob. AP1.18CPT
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