8) A closer examination of the records of the air samples in Example 3 reveals that each line of the data actually represents readings on the same day: 2.92 and 1.84 are from day 1, and so forth. Since this affects the validity of the results obtained in Example 10, reanalyze. (Use a=0.05.) Area A 2.92 1.88 5.35 3.81 4.69 4.86 5.81 5.55 Area B 1.84 0.95 4.26 3.18 3.44 3.69 4.95 4.47

MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Can someone please help? Thank you.
8) A closer examination of the records of the air samples in Example 3 reveals that each line of the data actually
represents readings on the same day: 2.92 and 1.84 are from day 1, and so forth. Since this affects the validity
of the results obtained in Example 10, reanalyze. (Use a=0.05.)
Area A
2.92
1.88
5.35
3.81
4.69
4.86
5.81
5.55
Area B
1.84
0.95
4.26
3.18
3.44
3.69
4.95
4.47
Transcribed Image Text:8) A closer examination of the records of the air samples in Example 3 reveals that each line of the data actually represents readings on the same day: 2.92 and 1.84 are from day 1, and so forth. Since this affects the validity of the results obtained in Example 10, reanalyze. (Use a=0.05.) Area A 2.92 1.88 5.35 3.81 4.69 4.86 5.81 5.55 Area B 1.84 0.95 4.26 3.18 3.44 3.69 4.95 4.47
Expert Solution
Step 1: Description of the steps of test :

By the given data,we use R- software, we get

Step :1

Hypothesis are 

Null hypothesis  H0: μ1=μ2

Alternative hypothesis  Ha : μ1μ2

Step : 2

Test statistic  t= 1.4779

Step :3

Degrees of freedom df= n1+n2-2=8+8-2=14

P value: 0.1616

Step :4

Decision: as p value > α=0.05, so fail to reject H0. 

Step :5

Conclusion: 

There is sufficient evidence to support the claim that the mean pollution indexes are the same for the two areas.


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