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 10.3, Problem 97E
To determine

To find from the following option a 95% confidence interval for μAB is given by which option.

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Answer to Problem 97E

Option (e) is the correct option.

Explanation of Solution

It is given that there are two varieties of tomato i.e. Variety A and Variety B. Then the researcher wants to claim that the variety A have a different average yield from that of variety B. we have to find out the confidence interval from the option.

Thus,

  n=10x¯AB=0.34sAB=0.83c=0.95

Thus the degree of freedom will be:

  df=n1=101=9

And the t -value will be:

  tα/2=2.262

Thus the confidence interval will be:

  x¯±tα/2×sn=0.34±2.262×0.8310

Thus, option (e) is the correct option.

Chapter 10 Solutions

PRACTICE OF STATISTICS F/AP EXAM

Ch. 10.1 - Prob. 11ECh. 10.1 - Prob. 12ECh. 10.1 - Prob. 13ECh. 10.1 - Prob. 14ECh. 10.1 - Prob. 15ECh. 10.1 - Prob. 16ECh. 10.1 - Prob. 17ECh. 10.1 - Prob. 18ECh. 10.1 - Prob. 19ECh. 10.1 - Prob. 20ECh. 10.1 - Prob. 21ECh. 10.1 - Prob. 22ECh. 10.1 - Prob. 23ECh. 10.1 - Prob. 24ECh. 10.1 - Prob. 25ECh. 10.1 - Prob. 26ECh. 10.1 - Prob. 27ECh. 10.1 - Prob. 28ECh. 10.1 - Prob. 29ECh. 10.1 - Prob. 30ECh. 10.1 - Prob. 31ECh. 10.1 - Prob. 32ECh. 10.1 - Prob. 33ECh. 10.1 - Prob. 34ECh. 10.1 - Prob. 35ECh. 10.1 - Prob. 36ECh. 10.2 - Prob. 37ECh. 10.2 - Prob. 38ECh. 10.2 - Prob. 39ECh. 10.2 - Prob. 40ECh. 10.2 - Prob. 41ECh. 10.2 - Prob. 42ECh. 10.2 - Prob. 43ECh. 10.2 - Prob. 44ECh. 10.2 - Prob. 45ECh. 10.2 - Prob. 46ECh. 10.2 - Prob. 47ECh. 10.2 - Prob. 48ECh. 10.2 - Prob. 49ECh. 10.2 - Prob. 50ECh. 10.2 - Prob. 51ECh. 10.2 - Prob. 52ECh. 10.2 - Prob. 53ECh. 10.2 - Prob. 54ECh. 10.2 - Prob. 55ECh. 10.2 - Prob. 56ECh. 10.2 - Prob. 57ECh. 10.2 - Prob. 58ECh. 10.2 - Prob. 59ECh. 10.2 - Prob. 60ECh. 10.2 - Prob. 61ECh. 10.2 - Prob. 62ECh. 10.2 - Prob. 63ECh. 10.2 - Prob. 64ECh. 10.2 - Prob. 65ECh. 10.2 - Prob. 66ECh. 10.2 - Prob. 67ECh. 10.2 - Prob. 68ECh. 10.2 - Prob. 69ECh. 10.2 - Prob. 70ECh. 10.2 - Prob. 71ECh. 10.2 - Prob. 72ECh. 10.2 - Prob. 73ECh. 10.2 - Prob. 74ECh. 10.3 - Prob. 75ECh. 10.3 - Prob. 76ECh. 10.3 - Prob. 77ECh. 10.3 - Prob. 78ECh. 10.3 - Prob. 79ECh. 10.3 - Prob. 80ECh. 10.3 - Prob. 81ECh. 10.3 - Prob. 82ECh. 10.3 - Prob. 83ECh. 10.3 - Prob. 84ECh. 10.3 - Prob. 85ECh. 10.3 - Prob. 86ECh. 10.3 - Prob. 87ECh. 10.3 - Prob. 88ECh. 10.3 - Prob. 89ECh. 10.3 - Prob. 90ECh. 10.3 - Prob. 91ECh. 10.3 - Prob. 92ECh. 10.3 - Prob. 93ECh. 10.3 - Prob. 94ECh. 10.3 - Prob. 95ECh. 10.3 - Prob. 96ECh. 10.3 - Prob. 97ECh. 10.3 - Prob. 98ECh. 10.3 - Prob. 99ECh. 10.3 - Prob. 100ECh. 10.3 - Prob. 101ECh. 10.3 - Prob. 102ECh. 10 - Prob. R10.1RECh. 10 - Prob. R10.2RECh. 10 - Prob. R10.3RECh. 10 - Prob. R10.4RECh. 10 - Prob. R10.5RECh. 10 - Prob. R10.6RECh. 10 - Prob. R10.7RECh. 10 - Prob. T10.1SPTCh. 10 - Prob. T10.2SPTCh. 10 - Prob. T10.3SPTCh. 10 - Prob. T10.4SPTCh. 10 - Prob. T10.5SPTCh. 10 - Prob. T10.6SPTCh. 10 - Prob. T10.7SPTCh. 10 - Prob. T10.8SPTCh. 10 - Prob. T10.9SPTCh. 10 - Prob. T10.10SPTCh. 10 - Prob. T10.11SPTCh. 10 - Prob. T10.12SPTCh. 10 - Prob. T10.13SPTCh. 10 - Prob. AP3.1CPTCh. 10 - Prob. AP3.2CPTCh. 10 - Prob. AP3.3CPTCh. 10 - Prob. AP3.4CPTCh. 10 - Prob. AP3.5CPTCh. 10 - Prob. AP3.6CPTCh. 10 - Prob. AP3.7CPTCh. 10 - Prob. AP3.8CPTCh. 10 - Prob. AP3.9CPTCh. 10 - Prob. AP3.10CPTCh. 10 - Prob. AP3.11CPTCh. 10 - Prob. AP3.12CPTCh. 10 - Prob. AP3.13CPTCh. 10 - Prob. AP3.14CPTCh. 10 - Prob. AP3.15CPTCh. 10 - Prob. AP3.16CPTCh. 10 - Prob. AP3.17CPTCh. 10 - Prob. AP3.18CPTCh. 10 - Prob. AP3.19CPTCh. 10 - Prob. AP3.20CPTCh. 10 - Prob. AP3.21CPTCh. 10 - Prob. AP3.22CPTCh. 10 - Prob. AP3.23CPTCh. 10 - Prob. AP3.24CPTCh. 10 - Prob. AP3.25CPTCh. 10 - Prob. AP3.26CPTCh. 10 - Prob. AP3.27CPTCh. 10 - Prob. AP3.28CPTCh. 10 - Prob. AP3.29CPTCh. 10 - Prob. AP3.30CPTCh. 10 - Prob. AP3.31CPTCh. 10 - Prob. AP3.32CPTCh. 10 - Prob. AP3.33CPTCh. 10 - Prob. AP3.34CPTCh. 10 - Prob. AP3.35CPT
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