A pharmacologist decided to test two common headache-tablets for their effective
ID: 3314983 • Letter: A
Question
A pharmacologist decided to test two common headache-tablets for their effectiveness: Tablet A 500mg and Tablet B 20 mg. The experiment was conducted as follows: A random patient that walked into the clinic and complained of a headache was given either tablet A, tablet B, or a placebo. After swallowing the tablet the patient was asked to stay in the clinic for an hour and afterwards to report whether the headache had disappeared, improved, or if the tablet had no effect (i.e. no improvement or even a worsening of the headache intensity). Use a 1% level of significance to test the claim that headache status is independent of headache relief tablet used.
Headache Disappeared
Headache Improved
No Change
Total
Tablet A
60
15
25
100
Tablet B
60
15
25
100
Placebo
60
35
15
110
Total
180
65
65
310
Contingency Table: Headache 7
Hypotheses:
H0: Headache status is independent of headache relief tablet used.
H1: Headache status is dependent on headache relief tablet used.
Expected Values:
Complete the 3x3 table of expected outcomes (round values to 3 decimal places).
Headache Disappeared
Headache Improved
No Change
Tablet A
20.968
Tablet B
20.968
Placebo
63.871
Results:
Calculate the test statistic (use two decimal places).
State the critical value from Table 3 in the textbook (page 242).
Conclusion:
We sufficient evidence to support the claim that patient headache status is dependent on which headache relief tablet was used (p 0.01).
(Use “have” or “lack” for the first blank and “<” or “>” for the second blank.)
Headache Disappeared
Headache Improved
No Change
Total
Tablet A
60
15
25
100
Tablet B
60
15
25
100
Placebo
60
35
15
110
Total
180
65
65
310
Explanation / Answer
We sufficient evidence to support the claim that patient headache status is dependent on which headache relief tablet was used (p 0.01)
Given table data is as below MATRIX col1 col2 col3 TOTALS row 1 60 15 25 100 row 2 60 15 25 100 row 3 60 35 15 110 TOTALS 180 65 65 N = 310 ------------------------------------------------------------------calculation formula for E table matrix E-TABLE col1 col2 col3 row 1 row1*col1/N row1*col2/N row1*col3/N row 2 row2*col1/N row2*col2/N row2*col3/N row 3 row3*col1/N row3*col2/N row3*col3/N ------------------------------------------------------------------
expected frequecies calculated by applying E - table matrix formulae E-TABLE col1 col2 col3 row 1 58.065 20.968 20.968 row 2 58.065 20.968 20.968 row 3 63.871 23.065 23.065 ------------------------------------------------------------------
calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 60 58.065 1.935 3.744 0.064 15 20.968 -5.968 35.617 1.699 25 20.968 4.032 16.257 0.775 60 58.065 1.935 3.744 0.064 15 20.968 -5.968 35.617 1.699 25 20.968 4.032 16.257 0.775 60 63.871 -3.871 14.985 0.235 35 23.065 11.935 142.444 6.176 15 23.065 -8.065 65.044 2.82 ^2 o = 14.307 ------------------------------------------------------------------
set up null vs alternative as
null, Ho: no relation b/w X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
level of significance, = 0.01
from standard normal table, chi square value at right tailed, ^2 /2 =13.277
since our test is right tailed,reject Ho when ^2 o > 13.277
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 14.307
critical value
the value of |^2 | at los 0.01 with d.f (r-1)(c-1)= ( 3 -1 ) * ( 3 - 1 ) = 2 * 2 = 4 is 13.277
we got | ^2| =14.307 & | ^2 | =13.277
make decision
hence value of | ^2 o | > | ^2 | and here we reject Ho
^2 p_value =0.006
ANSWERS
---------------
null, Ho: no relation b/w X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
test statistic: 14.307 =14.31
critical value: 13.277
p-value:0.006
decision: reject Ho
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