11(1.) Treating Bipolar Mania: Researchers conducted a randomized, double-blind
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Question
11(1.) Treating Bipolar Mania: Researchers conducted a randomized, double-blind study to measure the effects of the drug olanzapine on patients diagnosed with bipolar disorder. A total of 115 patients with a DSM-IV diagnosis of bipolar disorder were randomly divided into two groups. Group 1 ( n1 = 55 ) received 5 to 20 mg per day of olanzapine, while group 2 ( n2 = 60 ) received a placebo. The effectiveness of the drug was measured using the Young–Mania rating scale total score with the net improvement in the score recorded. The results are presented in the table. Does the evidence suggest that the experimental group experienced a larger mean improvement than the control group at the = 0.01 level of significance?
Experimental Group
Control Group
n
55
60
Mean improvement
14.8
8.1
Sample standard deviation
12.5
12.7
Source: “Efficacy of Olanzapine in Acute Bipolar Mania,” Archives of General Psychiatry, 59(9), pp. 841–848.
a. What is the null and alternative hypothesis?
b. Calculate the test statistic (Use 2-SampTTest on the TI Calculator and remember the test statistic is t)
Round 3 decimals
c. Calculate the degrees of freedom (you can use calculator or formula)
d. Look up the critical t-value for = 0.01 (Use T-Table)
e. Draw a normal curve that depicts the critical region. Label your test statistic and the critical value on the diagram.
f. Is the test statistic in the critical region?
g. What is the result of the test? Reject H0 or Do Not Reject H0?
h. State the conclusion.
Experimental Group
Control Group
n
55
60
Mean improvement
14.8
8.1
Sample standard deviation
12.5
12.7
Explanation / Answer
a)
Null hypothesis is drug having same effect as placebo
drug mean improvement = placebo mean improvment
Alternate hyptohesis is drug's improvement higher than placebo
drug mean improvement > placebo mean improvment
b)
t-stat for two groups
c)
degrees of freedom = n1+n2-2 = 50-65-2 = 113
d)
critical t-value for 0.01 alpha is -2.35 for single tailed test
f)
critical region is test stat greater than -2.35
since the test-stat is 2.37 it is very much in the critical region.
g)
since p-value is less than 0.01, reject null hypothesis.
h)
drug mean improvement > placebo mean improvment
drug has more improvemnt than placebo.
n 50 65 mean 14.8 8.1 sd 12.5 12.7 df 113 pooled sd sqrt( ((n1-1)*s1^2 +(n2-1)*s2^2)/(n1+n2-2) 12.61366 std error s * sqrt(1/n1+/n2) 2.372731 t-stat diff in means/se 2.82375 p-value 0.002806Related Questions
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