Question

Accupril is meant to control hypertension. In clinical trials of Accupril, 2142 subjects were divided into two groups. The 1563 subjects in the experimental group received Accupril. The 579 subjects in the control group received a placebo. Of the 1563 in the experimental group, 61 experienced dizziness as a side effect. Of the 579 subjects in the control group, 15 experienced dizziness as a side effect.
Let p_1 be the true proportion of people who experience dizziness while taking Accupril. Let p_2 be the true proportion of people who experience dizziness but do not take Accupril. Create a 95% confidence interval for p_1 - p_2.
a. (0.006, 0.092)
b. (-0.06, 0.92)
c. (-0.003, 0.029)
d. (-0.04, 0.29)

Answer 1

Answer:

Option C.

Step-by-step explanation:

Given information

$n_1=1563$ and $n_2=579$

$x_1=61$ and $x_2=15$

Using the given information we get

$p_1=\dfrac{x_1}{n_1}=\dfrac{61}{1563}\approx 0.039$

$p_2=\dfrac{x_2}{n_2}=\dfrac{15}{579}\approx 0.026$

The formula for confidence interval for p_1 - p_2 is

$C.I.=(p_1-p_2)\pm z*\sqrt{\dfrac{p_1(1-p_1)}{n_1}+\dfrac{p_2(1-p_2)}{n_2}}$

From the standard normal table the value of z* at 95% confidence interval = 1.96.

$C.I.=(0.039-0.026)\pm (1.96)\sqrt{\dfrac{0.039(1-0.039)}{1563}+\dfrac{0.026(1-0.026)}{579}}$

$C.I.=0.013\pm (1.96)(0.008)$

$C.I.=0.013\pm 0.016$

$C.I.=(0.013-0.016,0.013+0.016)$

$C.I.=(-0.003,0.029)$

The 95% confidence interval for p_1 - p_2. is (-0.003,0.029).

Therefore, the correct option is C.