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The contingency table shows the results of a random sample of former smokers by the number of times they tried to quit smoking before they were habit-free and gender. At a = 0.05, can you conclude that the number of times they tried to quit before they were habit-free is related to gender? Gender Male Female E1,2 = 273.680 1550 670 388 1038 = 250.443 = Number of times tried to quit before habit-free 4 or more 2-3 Next calculate the expected frequency for the two remaining cells in the first row, rounding to three decimal places. Gender Male Female 1 275 149 E1,3 252 136 1 273.680 150.320 670 226 1038 = 145.877 = 143 83 Use the same procedure to find the expected frequency for the remaining cells in the contingency table, rounding to three decimal places. Number of times tried to quit before habit-free 2-3 4 or more 250.443 137.557 145.877 80.123

The contingency table shows the results of a random sample of former smokers by the number of times they tried to quit smoking before they were habit-free and gender. At a = 0.05, can you conclude that the number of times they tried to quit before they were habit-free is related to gender? Gender Male Female E1,2 = 273.680 1550 670 388 1038 = 250.443 = Number of times tried to quit before habit-free 4 or more 2-3 Next calculate the expected frequency for the two remaining cells in the first row, rounding to three decimal places. Gender Male Female 1 275 149 E1,3 252 136 1 273.680 150.320 670 226 1038 = 145.877 = 143 83 Use the same procedure to find the expected frequency for the remaining cells in the contingency table, rounding to three decimal places. Number of times tried to quit before habit-free 2-3 4 or more 250.443 137.557 145.877 80.123

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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calculate the test statistic

tion list ia 2 stion 1 tion 2 tion 3 tion 4 - tion 5 tion 6 View an example | 4 parts remaining The contingency table shows the results of a random sample of former smokers by the number of times they tried to quit smoking before they were habit-free and gender. At a = 0.05, can you conclude that the number of times they tried to quit before they were habit-free is related to gender? Gender Male Female E1,2 = 273.680 670-388 1038 = 250.443 Print Next calculate the expected frequency for the two remaining cells in the first row, rounding to three decimal places. = Number of times tried to quit before habit-free 4 or more 1 2-3 275 149 Gender Male Female E1,3 252 136 = 1 273.680 150.320 670-226 1038 Use the same procedure to find the expected frequency for the remaining cells in the contingency table, rounding to three decimal places. 143 83 = 145.877 Number of times tried to quit before habit-free 2-3 4 or more 250.443 137.557 X 145.877 80.123 Continue 20 er of times the er of times the nce test by co Continue
Transcribed Image Text:tion list ia 2 stion 1 tion 2 tion 3 tion 4 - tion 5 tion 6 View an example | 4 parts remaining The contingency table shows the results of a random sample of former smokers by the number of times they tried to quit smoking before they were habit-free and gender. At a = 0.05, can you conclude that the number of times they tried to quit before they were habit-free is related to gender? Gender Male Female E1,2 = 273.680 670-388 1038 = 250.443 Print Next calculate the expected frequency for the two remaining cells in the first row, rounding to three decimal places. = Number of times tried to quit before habit-free 4 or more 1 2-3 275 149 Gender Male Female E1,3 252 136 = 1 273.680 150.320 670-226 1038 Use the same procedure to find the expected frequency for the remaining cells in the contingency table, rounding to three decimal places. 143 83 = 145.877 Number of times tried to quit before habit-free 2-3 4 or more 250.443 137.557 X 145.877 80.123 Continue 20 er of times the er of times the nce test by co Continue
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