Question

A researcher has developed a new drug designed to reduce blood pressure. In an experiment, 21 subjects were assigned randomly to the treatment group and received the new experimental drug. The other 23 subjects were assigned to the control group and received a standard, well‑known treatment. After a suitable period, the reduction in blood pressure for each subject was recorded. A summary of these data is: ¯ Treatment group (new drug) 21 23.48 8.01 Control group (old drug) 23 18.52 7.15 A 95% confidence interval for the difference in reduction of blood pressure between these two drugs, using Option 2 and the conservative method for degrees of freedom is:

Answer 1

Answer:

Step-by-step explanation:

Hello!

The objective of the research is to compare the newly designed drug to reduce blood pressure with the standard drug to test if the new one is more effective.

Two randomly selected groups of subjects where determined, one took the standard drug (1- Control) and the second one took the new drug (2-New)

1. Control

X₁: Reduction of the blood pressure of a subject that took the standard drug.

n₁= 23

X[bar]= 18.52

S= 7.15

2. New

X₂: Reduction of the blood pressure of a subject that took the newly designed drug.

n₂= 21

X[bar]₂= 23.48

S₂= 8.01

The parameter of study is the difference between the two population means (no order is specified, I'll use New-Standard) μ₂ - μ₁

Assuming both variables have a normal distribution, there are two options to estimate the difference between the two means using a 95% CI.

1) The population variances are unknown and equal:

[(X[bar]₂-X[bar]₁)±$t_{n_1+n_2-2;1-\alpha /2}$*($Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }$)]

$t_{n_1+n_2-2;1-\alpha /2}= t_{23+21-2;1-0.025}= t_{42;0.975}= 2.018$

$Sa=\sqrt{\frac{(n_1-1)*S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{22*7.15^2+20*8.01^2}{42} }= 7.57$

[23.48-18.52]±2.018*($7.57*\sqrt{\frac{1}{21} +\frac{1}{23} }$)]

[0.349; 9.571]

2) The population variables are unknown and different:

Welche's approximation:

[(X[bar]₂-X[bar]₁)±$t_{Dfw;1-\alpha /2}$*( $\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} }$)]

$Df_{w}= \frac{(\frac{S_1^2}{n_1} +\frac{S^2_2}{n_2} )^2}{\frac{(\frac{S_1^2}{n_1} )^2}{n_1-1}+ \frac{(\frac{S_2^2}{n_2} )^2}{n_2-1} } = \frac{(\frac{7.15^2}{23} +\frac{8.01^2_2}{21} )^2}{\frac{(\frac{7.15^2}{21} )^2}{20}+ \frac{(\frac{8.01^2}{23} )^2}{22} } = 42.85= 42$

$t_{Df_w;1-\alpha /2}= t_{42; 0.975}= 2.018$

[(23.48-18.52)±2.018$\sqrt{\frac{7.15^2}{23} +\frac{8.01^2}{21} }$]

[0.324; 9.596]

I hope this helps!