Patients (20) with high blood pressure were randomly assigned into two groups, a placebo group and a treatment group. The placebo group received conventional treatment while the treatment group received a new drug that is expected to lower blood pressure. After treatment for a couple of months, blood pressures of the patients were measured.
Determine if the new drug had a significant effect on high blood pressure.
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Blood Pressure (systolic)
Placebo Group New Drug Group
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90 71
95 79
67 69
120 98
89 91
92 85
100 89
83 75
79 78
85 80
The hypothesis to be tested for this case is,
“H_0:\mu_1=\mu_2”
“Against”
“H_1:\mu_1\lt\mu_2”
To perform this hypothesis test, we use the “p-value” method. The obtained “p-value” is then compared with the default “\alpha=0.05” since we are not given the level of significance.
Applying the 2 sample t-test in “R”, the test can be performed by keying in the following commands,
x=c(90,95,67,120,89,92,100,83,79,85)
y=c(71,79,69,98,91,85,89,75,78,80)
t. test(x, y)
Running these commands gives us the following output,
Welch Two Sample t-test
data: x and y
t = 1.6081, df = 15.569, p-value = 0.1279
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-2.730473 19.730473
sample estimates:
mean of x mean of y
90.0 81.5
The “p-value” from this output is divided by 2 since the test we are conducting is one sided.
Therefore, “p-value=0.1279/2=0.06395” and the null hypothesis is rejected if “p-value\lt\alpha” .
Since “p-value=0.06395\gt\alpha=0.05,” we fail to reject the null hypothesis and conclude that there is no sufficient evidence to show that the new drug had a significant effect on high blood pressure as compared to placebo at “\alpha=0.05” level of significance.
