Beginning Statistics, 2nd Edition
Beginning Statistics, 2nd Edition
2nd Edition
ISBN: 9781932628678
Author: Carolyn Warren; Kimberly Denley; Emily Atchley
Publisher: Hawkes Learning Systems
Question
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Chapter 9.3, Problem 11E
To determine

To construct:

And interpret a 99% confidence interval for the true mean difference between the weights to determine the mean amount of weight lost by doing on the diet.

Expert Solution & Answer
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Answer to Problem 11E

Solution:

The required confidence interval for the mean of paired differences is. (12.19,7.3476).

Explanation of Solution

Given Information:

A dietician wants to see how much weight a certain diet can help patients lose. Nine people agree to participate in her study. She weighs each patient at the beginning of the study and after 30 days of being on diet. The results are shown in table.

Patients’ Weights (in Pounds)
Starting Weight 180 176 152 230 183 214 250 197 201
Ending Weight 172 167 144 219 172 201 239 186 195

Formula Used:

The paired difference for any pair data values of the two dependent samples are given by

d=x2x1

Where x2 is a value of the second sample,

And x1 is a value of the first sample which is paired with x2.

The mean of the paired differences is calculated as,

d¯=din

Where di is the paired difference of the ith pair of values.

Also n is the total number of data values.

The sample standard deviation of the paired differences is calculated as,

sd=(did¯)n1

Where di is the paired difference of the ith pair of values,

d¯ is the mean of the pair differences,

And n is the total number of values given in the data.

For Margin of error we need critical value tα2,

E=(tα2)(sdn)

The confidence interval is calculated as,

(d¯±E)

Where the lower endpoint is d¯E and the upper endpoint is d¯+E.

Calculation:

Starting Weight x1 Ending Weight
x2
di (did¯) (did¯)2
180 172 8 1.77 3.1329
176 167 9 0.77 0.5929
152 144 8 1.77 3.1329
230 219 11 1.23 1.5129
183 172 11 1.23 1.5129
214 201 13 3.23 10.4329
250 239 11 1.23 1.5129
197 186 11 1.23 1.5129
201 195 6 3.77 14.2129
Sum 88 37.5561

The paired difference is calculated as,

di=x2x1

Substitute 180 for x1 and 172 for x2 in the above equation.

d1=172180=8

Proceed in the same manner to calculate di for the rest of the data and refer table for the rest of the di values calculated. The size of each sample is 9 that is n=9. Then the mean of the paired differences is calculated as,

d¯=din=(8)+(9)+......+(11)+(6)9=889=9.77

Substitute 8 for d1 and 9.77 for d¯ in (d1d¯).

(d1d¯)=(8(9.77))(d1d¯)=1.77

Square the both sides of above equation.

(d1d¯)2=(1.77)2(d1d¯)2=3.1329

Proceed in the same manner to calculate (did¯)2 for all the 1in for the rest data and refer table for the rest of the (did¯)2 values calculated. Then the value of (did¯)2 is calculated as,

(did¯)2=3.1329+0.5929+3.1329.....+14.2129=37.5561

The standard deviation of the paired difference is calculated as,

sd=(did¯)n1

Substitute 37.5561 for (did¯)2 and 9 for n in the above equation.

sd=37.55618=4.6945=2.166

It is given that c=0.99 and,

α=10.99=0.01

To calculate the margin of error we need tα2, the critical value. The value of α is given to be 0.01 and n is given to be 9 so the degree of freedom is,

n1=91=8

Then,

tα2=t0.012=3.355

The margin of error is calculated as,

E=(tα2)(sdn)

Substitute 3.355 for tα2, 2.166 for sd and 9 for n.

E=3.355×2.1669=2.42231

The confidence interval is,

(d¯±E)=(9.772.42231,9.77+2.42231)=(12.19,7.3476)

Interpretation:

Since both the end points are negative so this implies that researchers are 99% confident that the mean difference between the weights was negative and lies between 12.19 and 7.3476 and hence this implies that the weight of the patients has reduced by going on the diet.

Conclusion:

Thus the confidence interval for the mean of paired differences is. (12.19,7.3476).

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Chapter 9 Solutions

Beginning Statistics, 2nd Edition

Ch. 9.1 - Prob. 11ECh. 9.1 - Prob. 12ECh. 9.1 - Prob. 13ECh. 9.1 - Prob. 14ECh. 9.1 - Prob. 15ECh. 9.1 - Prob. 16ECh. 9.2 - Prob. 1ECh. 9.2 - Prob. 2ECh. 9.2 - Prob. 3ECh. 9.2 - Prob. 4ECh. 9.2 - Prob. 5ECh. 9.2 - Prob. 6ECh. 9.2 - Prob. 7ECh. 9.2 - Prob. 8ECh. 9.2 - Prob. 9ECh. 9.2 - Prob. 10ECh. 9.2 - Prob. 11ECh. 9.2 - Prob. 12ECh. 9.2 - Prob. 13ECh. 9.2 - Prob. 14ECh. 9.2 - Prob. 15ECh. 9.2 - Prob. 16ECh. 9.2 - Prob. 17ECh. 9.2 - Prob. 18ECh. 9.2 - Prob. 19ECh. 9.2 - Prob. 20ECh. 9.2 - Prob. 21ECh. 9.2 - Prob. 22ECh. 9.2 - Prob. 23ECh. 9.2 - Prob. 24ECh. 9.3 - Prob. 1ECh. 9.3 - Prob. 2ECh. 9.3 - Prob. 3ECh. 9.3 - Prob. 4ECh. 9.3 - Prob. 5ECh. 9.3 - Prob. 6ECh. 9.3 - Prob. 7ECh. 9.3 - Prob. 8ECh. 9.3 - Prob. 9ECh. 9.3 - Prob. 10ECh. 9.3 - Prob. 11ECh. 9.3 - Prob. 12ECh. 9.3 - Prob. 13ECh. 9.3 - Prob. 14ECh. 9.3 - Prob. 15ECh. 9.3 - Prob. 16ECh. 9.3 - Prob. 17ECh. 9.3 - Prob. 18ECh. 9.3 - Prob. 19ECh. 9.3 - Prob. 20ECh. 9.3 - Prob. 21ECh. 9.4 - Prob. 1ECh. 9.4 - Prob. 2ECh. 9.4 - Prob. 3ECh. 9.4 - Prob. 4ECh. 9.4 - Prob. 5ECh. 9.4 - Prob. 6ECh. 9.4 - Prob. 7ECh. 9.4 - Prob. 8ECh. 9.4 - Prob. 9ECh. 9.4 - Prob. 10ECh. 9.4 - Prob. 11ECh. 9.4 - Prob. 12ECh. 9.4 - Prob. 13ECh. 9.4 - Prob. 14ECh. 9.4 - Prob. 15ECh. 9.4 - Prob. 16ECh. 9.4 - Prob. 17ECh. 9.4 - Prob. 18ECh. 9.4 - Prob. 19ECh. 9.4 - Prob. 20ECh. 9.4 - Prob. 21ECh. 9.4 - Prob. 22ECh. 9.4 - Prob. 23ECh. 9.4 - Prob. 24ECh. 9.4 - Prob. 25ECh. 9.5 - Prob. 1ECh. 9.5 - Prob. 2ECh. 9.5 - Prob. 3ECh. 9.5 - Prob. 4ECh. 9.5 - Prob. 5ECh. 9.5 - Prob. 6ECh. 9.5 - Prob. 7ECh. 9.5 - Prob. 8ECh. 9.5 - Prob. 9ECh. 9.5 - Prob. 10ECh. 9.5 - Prob. 11ECh. 9.5 - Prob. 12ECh. 9.5 - Prob. 13ECh. 9.5 - Prob. 14ECh. 9.5 - Prob. 15ECh. 9.5 - Prob. 16ECh. 9.5 - Prob. 17ECh. 9.5 - Prob. 18ECh. 9.5 - Prob. 19ECh. 9.5 - Prob. 20ECh. 9.5 - Prob. 21ECh. 9.5 - Prob. 22ECh. 9.5 - Prob. 23ECh. 9.5 - Prob. 24ECh. 9.5 - Prob. 25ECh. 9.5 - Prob. 26ECh. 9.5 - Prob. 27ECh. 9.5 - Prob. 28ECh. 9.CR - Prob. 1CRCh. 9.CR - Prob. 2CRCh. 9.CR - Prob. 3CRCh. 9.CR - Prob. 4CRCh. 9.CR - Prob. 5CRCh. 9.CR - Prob. 6CRCh. 9.CR - Prob. 7CRCh. 9.CR - Prob. 8CRCh. 9.CR - Prob. 9CRCh. 9.CR - Prob. 10CRCh. 9.CR - Prob. 11CRCh. 9.CR - Prob. 12CRCh. 9.CR - Prob. 13CRCh. 9.CR - Prob. 14CRCh. 9.CR - Prob. 15CRCh. 9.P - Prob. 1P
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