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.CR, Problem 12CR
To determine

Construct and interpret a confidence interval for the true difference between the two population means.

Expert Solution & Answer
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Answer to Problem 12CR

Solution:

The confidence interval is (17.9171,31.5029) and the length timing from Atlanta to Dallas is more than the length timing from Dallas to Atlanta so the flight from Dallas to Atlanta is faster.

Explanation of Solution

Given Information:

A business person wants to determine how much faster the return trip from Dallas to Atlanta is than the initial trip from Atlanta to Dallas. He records the times for his next seven flights to and from Dallas and obtains the following data. Assuming that both population distributions arc approximately normal and the variances of the two populations are not the same.

Atlanta to Dallas Dallas to Atlanta
1 h 49 min 1 h 32 min
1 h 51 min 1 h 23 min
2 h 03 min 1 h 38 min
1 h 47 min 1 h 26 min
1 h 53 min 1 h 30 min
2 h 01 min 1 h 19 min
1 h 45 min 1 h 28 min

Formula used:

When the standard deviations of both the populations are unknown and assumed to be unequal, the samples taken are independent of each other and the population distribution are approximately normal, then the margin of error is calculates as,

E=tα/2s12n1+s22n2

Where s1 and s2 are the two sample standard deviations,

And n1 and n2 are two sample sizes.

The number of degrees of freedom for the t-distribution is given by the smaller of the values n11 and n21.

Then, the confidence interval for the difference between two population means for independent data sets is given by

(x¯1 - x¯2)- E < μ1 - μ2 < (x¯1-x¯2)+ E

or

((x¯1 - x¯2)- E,(x¯1 - x¯2)+ E)

Where x¯1 and x¯2 are the two sample means.

Calculation:

Atlanta to Dallas Dallas to Atlanta
109 min 92 min
111 min 83 min
123 min 98 min
107 min 86 min
113 min 90 min
121 min 79 min
105 min 88 min

Let the length of time the flight takes from Atlanta to Dallas be sample 1 and the length of time the flight takes from Dallas to Atlanta be sample 2.

For the population whose population variances are unequal, the t –distribution is applied for the minimum degree of freedom .i.e. the minimum of n1-1 and n2-1.

n11=71=6n21=71=6

Hence the degree of freedom is 6.

It is given that the level of confidence is 0.90 then the level of significance is,

α=10.90=0.10

Then,

tα/2=t0.10/2=t0.05=1.943

The mean is calculated as,

x¯=xin

Where xi is the data at the ith position,

And n is the total number of observations.

Th mean of the sample 1 is calculated as,

x¯1=109+111+123+107+113+121+1057=7897=112.71

The standard deviation is calculated as,

sd2=(xix¯)n1

The standard deviation of sample 1 be s12 calculated as,

s12=(xix¯1)n11

Substitute 112.71 for x¯1, 7 for n1 and the respective xi values.

s12=(109112.71)2+(111112.71)2+....+(105112.71)271=283.426=47.23

The mean of the sample 2 is calculated as,

x¯2=92+83+98+86+90+79+887=88

The standard deviation of sample 2 be s22 calculated as,

s22=(xix¯2)n21

Substitute 88 for x¯2, 7 for n2 and the respective xi values.

s22=(9288)2+(8388)2+(9888)2+(8688)2+..+(8888)271=2306=38.33

The margin of error is calculated as,

E=tα/2s12n1+s22n2

Substitute t0.05=1.943,s12=47.23,n1=7,s22=38.33,n2=7 in the above formula.

E=1.943×47.237+38.337=6.7929

The confidence interval is,

((x¯1 - x¯2)- E,(x¯1 - x¯2)+ E)

Substitute 112.71 for x¯1, 88 for x¯2 and 6.7929 for E in the above formula ((x¯1 - x¯2)- E,(x¯1 - x¯2)+ E)=(((112.7188)6.7929),((112.7188)+6.7929))=(17.9171,31.5029)

Conclusion:

Both the end points are coming out to be positive that is the difference of the mean is also positive that is the length timing from Atlanta to Dallas is more than the length timing from Dallas to Atlanta so the flight from Dallas to Atlanta is faster.

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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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