In: Statistics and Probability
In the book Business Research Methods (5th ed.), Donald R.
Cooper and C. William Emory discuss studying the relationship
between on-the-job accidents and smoking. Cooper and Emory describe
the study as follows:
Suppose a manager implementing a smoke-free workplace policy is interested in whether smoking affects worker accidents. Since the company has complete reports of on-the-job accidents, she draws a sample of names of workers who were involved in accidents during the last year. A similar sample from among workers who had no reported accidents in the last year is drawn. She interviews members of both groups to determine if they are smokers or not.
The sample results are given in the following table.
On-the-Job Accident | ||||||
Smoker | Yes | No | Row Total | |||
Heavy | 12 | 8 | 20 | |||
Moderate | 9 | 10 | 19 | |||
Nonsmoker | 13 | 14 | 27 | |||
Column total | 34 | 32 | 66 | |||
Expected counts are below observed counts | ||||||
Accident | No Accident | Total | ||||
Heavy | 12 | 8 | 20 | |||
10.30 | 9.70 | |||||
Moderate | 9 | 10 | 19 | |||
9.79 | 9.21 | |||||
Nonsmoker | 13 | 14 | 27 | |||
13.91 | 13.09 | |||||
Total | 34 | 32 | 66 | |||
Chi-Sq = .83, DF = 2, P-Value = 0.660 | ||||||
(a) For each row and column total in the above
table, find the corresponding row/column percentage. (Round
your answers to 2 decimal places.)
Row 1 | % |
Row 2 | % |
Row 3 | % |
Column 1 | % |
Column 2 | % |
(b) For each cell in the above table, find the
corresponding cell, row, and column percentages. (Round
your answers to 2 decimal places.)
Accident | No Accident | ||
Heavy | Cell= % | Cell= % | |
Row= % | Row= % | ||
Column= % | Column= % | ||
Moderate | Cell= % | Cell= % | |
Row= % | Row= % | ||
Column= % | Column= % | ||
Nonsmoker | Cell= % | Cell= % | |
Row= % | Row= % | ||
Column= % | Column= % | ||
(c) Use the MINITAB output in the above to test
the hypothesis that the incidence of on-the-job accidents is
independent of smoking habits. Set α = .01.
(Click to select)Do not rejectReject
H0.
(d) Is there a difference in on-the-job
accident occurrences between smokers and nonsmokers?
Conclude there is (Click to select)no differencedifference between smokers and nonsmokers.
Solution:-
A)
Percentage | |
Row 1 | 30.30% |
Row 2 | 28.79% |
Row 3 | 40.91% |
Column 1 | 51.52% |
Column 2 | 48.48% |
b)
Accident | No Accident | ||
Heavy | Cell = 18.18% | Cell = 12.12% | |
Row = 60 % | Row = 40% | ||
Column = 35.29 % | Column = 25% | ||
Moderate | Cell = 13.64% | Cell = 15.15% | |
Row = 47.37% | Row = 52.63% | ||
Column = 26.47% | Column = 31.25 % | ||
Nonsmoker | Cell = 19.70% | Cell = 21.21% | |
Row = 48.15% | Row = 51.85% | ||
Column = 38.24 % | Column = 43.75 % | ||
c)
Smoker |
Yes |
Expected | [(Or,c - Er,c)2/ Er,c] | No | Expected | [(Or,c - Er,c)2/ Er,c] | Total |
Heavy | 12 | 10 | 0.279500891 | 8 | 9.696969697 | 0.296969697 | 20 |
Moderate | 9 | 10 | 0.063420584 | 10 | 9.212121212 | 0.06738437 | 19 |
Nonsmoker | 13 | 14 | 0.059417706 | 14 | 13.09090909 | 0.063131313 | 27 |
Total | 34 | 34 | 0.40 | 32.00 | 32.00 | 0.43 | 66 |
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
H0: Incidence of on-the-job accidents is
independent of smoking habits
Ha: Incidence of on-the-job accidents is not
independent of smoking habits.
Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a chi-square test for independence.
Analyze sample data. Applying the chi-square test for independence to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.
1) The sampling method is simple random sampling.
2) The variables under study are each categorical.
3) If sample data are displayed in a contingency table, the expected frequency count for each cell of the table is at least 5.
DF = (r - 1) * (c - 1) = (2 - 1) * (3 - 1)
D.F = 2
Er,c = (nr * nc) / n
Χ2 = 0.83
where DF is the degrees of freedom.
The P-value is the probability that a chi-square statistic having 2 degrees of freedom is more extreme than 0.82.
We use the Chi-Square Distribution Calculator to find P(Χ2 > 0.83) = 0.66
Interpret results. Since the P-value (0.66) is greater than the significance level (0.01), we have to accept the null hypothesis.
Thus, we conclude that the incidence of on-the-job accidents is independent of smoking habits.
d) No, there is no difference in on-the-job accident occurrences between smokers and nonsmokers.