Researchers selected a simple random sample of 4048 medical records of adults diagnosed with gum disease....

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Researchers selected a simple random sample of 4048 medicalrecords of adults diagnosed with gum disease. In all, 2226 werecurrent smokers, 891 were former smokers, and 931 never smokedregularly. Their research question is:

Do these data indicate that gum disease is equally likelyregardless of smoking status?

Using a significance level of 0.05, what is the appropriateconclusion for this test?

The data are consistent with an equal representation of current,former, and never smokers among adults diagnosed with gumdisease.

Current smokers make up a significantly greater proportion ofadults diagnosed with gum disease than former or never smokers.

Current smokers are most likely to have gum disease.

There is significant evidence that current, former, and neversmokers are not equally represented among adults diagnosed with gumdisease.

Answer & Explanation Solved by verified expert
3.5 Ratings (269 Votes)

Chi square test for Goodness of fit  
  
expected frequncy,E = expected proportions*total frequency  
total frequency=   4048

observed frequencey, O expected proportion expected frequency,E (fi - ei) (O-E)²/E
2226 0.333 1349.333 876.667 569.573
891 0.333 1349.333 -458.333 155.684
931 0.333 1349.333 -418.333 129.696

chi square test statistic,X² = Σ(O-E)²/E =   854.9531              
                  
level of significance, α=   0.05              
Degree of freedom=k-1=   3   -   1   =   2
                  
P value =   0.0000   [ excel function: =chisq.dist.rt(test-stat,df) ]          
Decision: P value < α, Reject Ho                  

There is significant evidence that current, former, and never smokers are not equally represented among adults diagnosed with gum disease.


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