The Practice of Statistics for AP - 4th Edition
The Practice of Statistics for AP - 4th Edition
4th Edition
ISBN: 9781429245593
Author: Starnes, Daren S., Yates, Daniel S., Moore, David S.
Publisher: Macmillan Higher Education
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Chapter 8.3, Problem 66E
To determine

To calculate: The critical value t* when the confidence level is 90%.

Expert Solution & Answer
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Answer to Problem 66E

The critical value is 1.671, with technology critical value is 1.665 and the advantage for more accurate is that 90% of the confidence interval contains the true population mean.

Explanation of Solution

Given Information:

Sample size n=77

Confidence interval c=90%

Formula Used Degree of freedom df=n1

Calculation:

Find the degree of freedom, use the formula df=n1 .

  df=n1=771=76

Table B does not contain a row with df=76 .Use the row that is closest to df=76 .

Thus, df=60

The confidence level is 90%.

Convert 90% into decimal.

  90100=0.90

Find the value of the column.

  1c2=10.902=0.05

From table B find critical value t* , use the row and column.

Thus, the critical value t*is1.671 .

Find the critical value, use technology for df=76 and 2p(X>x)=0.01 .

Thus, critical value t*is1.665

Thus, the critical value is 1.671 and with technology critical value is 1.665.

Higher the accuracy of degree of freedom more accurate will be the confidence interval. Thus, 90% of the confidence interval contains the true population mean for the sample size of 77.

Hence, 90% of the confidence interval contains the true population mean.

Chapter 8 Solutions

The Practice of Statistics for AP - 4th Edition

Ch. 8.1 - Prob. 8ECh. 8.1 - Prob. 9ECh. 8.1 - Prob. 10ECh. 8.1 - Prob. 11ECh. 8.1 - Prob. 12ECh. 8.1 - Prob. 13ECh. 8.1 - Prob. 14ECh. 8.1 - Prob. 15ECh. 8.1 - Prob. 16ECh. 8.1 - Prob. 17ECh. 8.1 - Prob. 18ECh. 8.1 - Prob. 19ECh. 8.1 - Prob. 20ECh. 8.1 - Prob. 21ECh. 8.1 - Prob. 22ECh. 8.1 - Prob. 23ECh. 8.1 - Prob. 24ECh. 8.1 - Prob. 25ECh. 8.1 - Prob. 26ECh. 8.2 - Prob. 1.1CYUCh. 8.2 - Prob. 1.2CYUCh. 8.2 - Prob. 2.1CYUCh. 8.2 - Prob. 2.2CYUCh. 8.2 - Prob. 2.3CYUCh. 8.2 - Prob. 2.4CYUCh. 8.2 - Prob. 3.1CYUCh. 8.2 - Prob. 3.2CYUCh. 8.2 - Prob. 27ECh. 8.2 - Prob. 28ECh. 8.2 - Prob. 29ECh. 8.2 - Prob. 30ECh. 8.2 - Prob. 31ECh. 8.2 - Prob. 32ECh. 8.2 - Prob. 33ECh. 8.2 - Prob. 34ECh. 8.2 - Prob. 35ECh. 8.2 - Prob. 36ECh. 8.2 - Prob. 37ECh. 8.2 - Prob. 38ECh. 8.2 - Prob. 39ECh. 8.2 - Prob. 40ECh. 8.2 - Prob. 41ECh. 8.2 - Prob. 42ECh. 8.2 - Prob. 43ECh. 8.2 - Prob. 44ECh. 8.2 - Prob. 45ECh. 8.2 - Prob. 46ECh. 8.2 - Prob. 47ECh. 8.2 - Prob. 48ECh. 8.2 - Prob. 49ECh. 8.2 - Prob. 50ECh. 8.2 - Prob. 51ECh. 8.2 - Prob. 52ECh. 8.2 - Prob. 53ECh. 8.2 - Prob. 54ECh. 8.3 - Prob. 1.1CYUCh. 8.3 - Prob. 2.1CYUCh. 8.3 - Prob. 3.1CYUCh. 8.3 - Prob. 3.2CYUCh. 8.3 - Prob. 3.3CYUCh. 8.3 - Prob. 3.4CYUCh. 8.3 - Prob. 55ECh. 8.3 - Prob. 56ECh. 8.3 - Prob. 57ECh. 8.3 - Prob. 58ECh. 8.3 - Prob. 59ECh. 8.3 - Prob. 60ECh. 8.3 - Prob. 61ECh. 8.3 - Prob. 62ECh. 8.3 - Prob. 63ECh. 8.3 - Prob. 64ECh. 8.3 - Prob. 65ECh. 8.3 - Prob. 66ECh. 8.3 - Prob. 67ECh. 8.3 - Prob. 68ECh. 8.3 - Prob. 69ECh. 8.3 - Prob. 70ECh. 8.3 - Prob. 71ECh. 8.3 - Prob. 72ECh. 8.3 - Prob. 73ECh. 8.3 - Prob. 74ECh. 8.3 - Prob. 75ECh. 8.3 - Prob. 76ECh. 8.3 - Prob. 77ECh. 8.3 - Prob. 78ECh. 8.3 - Prob. 79ECh. 8.3 - Prob. 80ECh. 8 - Prob. 1CRECh. 8 - Prob. 2CRECh. 8 - Prob. 3CRECh. 8 - Prob. 4CRECh. 8 - Prob. 5CRECh. 8 - Prob. 6CRECh. 8 - Prob. 7CRECh. 8 - Prob. 8CRECh. 8 - Prob. 9CRECh. 8 - Prob. 10CRECh. 8 - Prob. 1PTCh. 8 - Prob. 2PTCh. 8 - Prob. 3PTCh. 8 - Prob. 4PTCh. 8 - Prob. 5PTCh. 8 - Prob. 6PTCh. 8 - Prob. 7PTCh. 8 - Prob. 8PTCh. 8 - Prob. 9PTCh. 8 - Prob. 10PTCh. 8 - Prob. 11PTCh. 8 - Prob. 12PTCh. 8 - Prob. 13PT
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