The table below shows the raw data for the results of a nicotine patch drug tria

Question

The table below shows the raw data for the results of a nicotine patch drug trial. The treatment column describes whether the participant received a placebo or the patch, the symptom column describes if the participant described relief from their nicotine cravings or not. a. Create a two-way table for the data. b. Write the null and alternate hypothesis for this test. c. What are the assumptions of this test? d. Conduct the appropriate test. Show all your work including the work for calculating one expected count and all parts of the statistic. e. Write a conclusion for this test (include your check of the assumptions).

Explanation / Answer

a)

b)

H0: Variable treatment and Variable symptom are independent.
Ha: Variable treatment and Variable symptom are not independent.

c) The sampling method is simple random sampling.
The variables under study are each categorical.
The expected frequency count for each cell of the table is at least 5.

d) Degrees of freedom. The degrees of freedom (DF) is equal to:
DF = (r - 1) * (c - 1) = (2-1)*(2-1) = 1
where r is the number of levels for one catagorical variable, and c is the number of levels for the other categorical variable.

Expected frequencies.
Er,c = (nr * nc) / n
where Er,c is the expected frequency count for level r of Variable A and level c of Variable B, nris the total number of sample observations at level r of Variable A, nc is the total number of sample observations at level c of Variable B, and n is the total sample size.

Expected count for Nicotine patch and no relief = (13*13)/25 = 6.76

Test statistic. The test statistic is a chi-square random variable (2) defined by the following equation.
2 = [ (Or,c - Er,c)2 / Er,c ]
where Or,c is the observed frequency count at level r of Variable A and level c of Variable B, and Er,c is the expected frequency count at level r of Variable A and level c of Variable B.
2 = [(4-6.76)2/6.76] + [(9-6.24)2/6.24] + [(9-6.24)2/6.24] + [(3-5.76)2/5.76] = 4.8909

P = 0.027

Yates chi-square, corrected for continuity.
2 = [(abs(4-6.76)-0.5)2/6.76] + [(9-6.24-0.5)2/6.24] + [(9-6.24-0.5)2/6.24] + [(abs(3-5.76)-0.5)2/5.76]= 3.28,
p = 0.0701

e) At = 0.1, The results are significant with p = 0.0701 < 0.1. There is sufficient evidence to reject the null hypothesis and suggest that relief is dependent on the treatment.

H0: Variable treatment and Variable symptom are independent.
Ha: Variable treatment and Variable symptom are not independent.

c) The sampling method is simple random sampling.
The variables under study are each categorical.
The expected frequency count for each cell of the table is at least 5.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.