Question

A certain naturopathic remedy being tested may be effective in increasing the life expectancy in people who have a rare disease. A control group consists of 25 subjects who have not taken the remedy and have a mean life expectancy of 72.1 years. A treatment group of the same size have been given the remedy have a mean life expectancy of 75.2 years. After re-randomizing through simulation the distribution plot for the differences of means is shown.
What is the Margin of Error, rounded to the nearest hundredth, and is the naturopathic remedy effective in increasing the life expectancy of people with this rare disease?  

A. M.E.= 5.74 and no, the remedy is not effective in increasing the life expectancy in people because the original difference of means of 3.1 years is statistically within what we would expect if the differences were random 

B. M.E.= 2.87 and yes, the remedy is effective in increasing the life expectancy in people because the original difference of means of 3.1 years is statistically beyond what we would expect if the differences were random 

C. M.E.= 2.87 and no, the remedy is not effective in increasing the life expectancy in people because the original difference of means of 3.1 years is statistically within what we would expect if the differences were random 

D. M.E.= 9.09 and yes, the remedy is effective in increasing the life expectancy in people because the original difference of means of 3.1 years is statistically beyond what we would expect if the differences were random

Answer 1

Answer:

Confidence Interval for Two Independent Samples

Step-by-step explanation:

First of all we define the two groups:

Group A with an average of 72.1

Group B with an average of 75.2

The general formula for the margin of error for a sample proportion (if certain conditions are met) is

$M.E = z* \sqrt{\frac{\rho'(1-\rho')}{n} }$ (1)

where

$\rho'$

is the sample proportion, n is the sample size, and z* is the appropriate z*-value for your desired level of confidence (from the following table).

Since our sample is 50, and our ideal group is 25, we get $\rho'$

$\rho'= \frac{25}{50} = 0.5$

Our sample size is

$n=50$

The Percentage Confidence (z) for a value of 95% is equivalent to 1.96 (This is possible to consult in any standardized table)

We replace

$M.E = 1.96* \sqrt{\frac{0.5(1-0.5)}{50} }$

$M.E= 1.96* (1.465)$

$M.E = 2.87$

Since the value of the error remains below the difference of the averages that is 3.1 (the subtraction of the two averages) then it is inferred that it is effective in increasing the life expectancy in people because the original difference of means of 3.1 years is statistically beyond what we would expect if the differences were random

Answer 2

The Margin of Error M.E.= 2.87 and yes, the remedy is effective in increasing the life expectancy in people because the original difference of means of 3.1 years is statistically beyond what we would expect if the differences were random.