Answer to Question #252530 in Statistics and Probability for Jude

Question

Ten patients with anorexia in the final stages of recovery were candidates for a trial of a new anti-anxiety drug. Patients received the drug over four weeks. Each patient was also given a placebo over four weeks. Anxiety scores were measured at the

end of each treatment period. The order of administration of the active drug and the placebo was randomized for each patient. There was a washout period of four weeks between the administration of the drug and placebo.

Was the new drug significantly more effective than the placebo in reducing patient anxiety?

Answer 1

Let $\mu_1$ be the mean for population 1(Drug) and $\mu_2$ be the mean for population 2(Placebo).

The hypotheses we test are,

$H_0:\mu_1=\mu_2$

$Against$

$H_1:\mu_1\gt\mu_2$

To perform this test, we shall use the $p-value$ technique then compare this value with the given $\alpha$ level of significance.

By applying the 2 sample t-test in $R$, we can easily perform the test using the commands below,

x=c(19,11,14,17,23,11,15,19,11,8)

y=c(22,18,17,19,22,12,14,11,19,7)

t. test(x, y)

These commands produce the following output,

  Welch Two Sample t-test

data: x and y

t = -0.60286, df = 17.944, p-value = 0.5541

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

 -5.831405 3.231405

sample estimates:

mean of x mean of y 

   14.8   16.1 

From this output, the $p-value$ is divided by 2 since we are conducting a one sided test.

Thus, $p-value=0.5541/2=0.27705$ and the null hypothesis is rejected if $p-value\lt\alpha$

Since $p-value=0.27705\gt\alpha=0.05$, we fail to reject the null hypothesis and conclude that there is no sufficient evidence to show that the new drug is significantly more effective than placebo in reducing patient anxiety at $\alpha=5\%$ level of significance.