Question #249331
A research group wishes to conduct a randomized trial to compare a new medication to a medication
they consider standard care. One hundred patients with hypertension are enrolled and randomized
to one of the two comparison treatments. After taking the assigned medication for 6 weeks, their
systolic blood pressure (SBP) is measured. Summary statistics are given here. Use the data to test
if there is a significant difference in systolic blood pressures between medication groups assuming
equality of the variances. Run the appropriate test at a 5% level of significance.
Treatment Number of Patients Mean SBP Standard Deviation
New 50 130 12
Standard 50 135 10
1
Expert's answer
2021-10-15T11:42:27-0400

n1=50x1ˉ=130s1=12n2=50x2ˉ=135s2=10H0:μ1=μ2H1:μ1μ2n_1= 50 \\ \bar{x_1} = 130 \\ s_1 = 12 \\ n_2 = 50 \\ \bar{x_2} = 135 \\ s_2 = 10 \\ H_0: \mu_1=\mu_2 \\ H_1: \mu_1 ≠ \mu_2

Test-statistic:

t=x1ˉx2ˉsp2×(1n1+1n2)sp2=(n11)×s12+(n21)×s22n1+n22=(501)×122+(501)×10250+502=122t=130135122×(150+150)=2.263t = \frac{\bar{x_1} - \bar{x_2}}{\sqrt{s^2_{p} \times (\frac{1}{n_1} + \frac{1}{n_2})}} \\ s^2_{p} = \frac{(n_1-1) \times s^2_1 + (n_2-1) \times s^2_2}{n_1+n_2-2} \\ = \frac{(50-1) \times 12^2 + (50-1) \times 10^2}{50+50-2} \\ = 122 \\ t = \frac{130-135}{\sqrt{122 \times (\frac{1}{50} + \frac{1}{50})}} \\ = -2.263

P-value

By using t distribution p-value table at α= 0.05, t = -2.263, df=n1+n22=98df = n_1+n_2- 2 = 98

p-value = 0.0258

p-value < level of significance

We have to reject H0 at α= 0.05.

There is sufficient evidence to conclude that the significant difference in systolic blood pressures between medication groups assuming equality of the variances.


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