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Refer to the table from the previous problem (provided below) to answer the question. West East North Total Non-Defective 93 1070 1208 2371 Defective 13 56 60 129 Total 106 1126 1268 2500 Calculate the test statistic and provide the corresponding P-value for this table. Then use the a = 0.05 significance level and your answer to form a conclusion regarding the null hypothesis. O A. x2 = 11.49, P-value = 0.0032. Therefore the null hypothesis is rejected, (P-value is less than 0.05). O B. 2 = 11.49, P-value = 0.0032. Therefore the null hypothesis is not rejected, (P-value is less than 0.05). O C. y2 = 11.49, P-value = 0.0016. Therefore the null hypothesis is rejected, (P-value is less than 0.05). O D. x2 = 11.49, P-value = 1.6032. Therefore the null hypothesis is not rejected, (P-value is less than 0.05).

Refer to the table from the previous problem (provided below) to answer the question. West East North Total Non-Defective 93 1070 1208 2371 Defective 13 56 60 129 Total 106 1126 1268 2500 Calculate the test statistic and provide the corresponding P-value for this table. Then use the a = 0.05 significance level and your answer to form a conclusion regarding the null hypothesis. O A. x2 = 11.49, P-value = 0.0032. Therefore the null hypothesis is rejected, (P-value is less than 0.05). O B. 2 = 11.49, P-value = 0.0032. Therefore the null hypothesis is not rejected, (P-value is less than 0.05). O C. y2 = 11.49, P-value = 0.0016. Therefore the null hypothesis is rejected, (P-value is less than 0.05). O D. x2 = 11.49, P-value = 1.6032. Therefore the null hypothesis is not rejected, (P-value is less than 0.05).

MATLAB: An Introduction with Applications
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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Refer to the table from the previous problem (provided below) to answer the question. West East North Total Non-Defective 93 1070 1208 2371 Defective 13 56 60 129 Total 106 1126 1268 2500 Calculate the test statistic and provide the corresponding P-value for this table. Then use the a = 0.05 significance level and your answer to form a conclusion regarding the null hypothesis. O A. x2 = 11.49, P-value = 0.0032. Therefore the null hypothesis is rejected, (P-value is less than 0.05). O B. x = 11.49, P-value = 0.0032. Therefore the null hypothesis is not rejected, (P-value is less than 0.05). O C. x2 = 11.49, P-value = 0.0016. Therefore the null hypothesis is rejected, (P-value is less than 0.05). O D. X = 11.49, P-value = 1.6032. Therefore the null hypothesis is not rejected, (P-value is less than 0.05).
Transcribed Image Text:Refer to the table from the previous problem (provided below) to answer the question. West East North Total Non-Defective 93 1070 1208 2371 Defective 13 56 60 129 Total 106 1126 1268 2500 Calculate the test statistic and provide the corresponding P-value for this table. Then use the a = 0.05 significance level and your answer to form a conclusion regarding the null hypothesis. O A. x2 = 11.49, P-value = 0.0032. Therefore the null hypothesis is rejected, (P-value is less than 0.05). O B. x = 11.49, P-value = 0.0032. Therefore the null hypothesis is not rejected, (P-value is less than 0.05). O C. x2 = 11.49, P-value = 0.0016. Therefore the null hypothesis is rejected, (P-value is less than 0.05). O D. X = 11.49, P-value = 1.6032. Therefore the null hypothesis is not rejected, (P-value is less than 0.05).
Expert Solution
Step 1

The expected value of a cell(E) = Row sub total×Column Sub totalGrand Total

_____________________________________________________________

E(Defective, West) = 129×1062500=5.5

similarly for others

_____________________________________________________________

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