Chapter PV, Problem 27RE

(a)

To determine

To create a $95\%$ confidence interval for the effectiveness of this drug.

Answer to Problem 27RE

  $(0.54429,0.62499)$ .

Explanation of Solution

It is given in the question that medical journal reported patients suffering from Crohn’s disease responded positively to injections of the drug. Thus, let us check the conditions and assumptions necessary for inference as:

Random condition: It is not satisfied as it is assumed that the sample is randomly selected.

Independent condition: It is satisfied as patients respond to the injections independently.

  $10\%$ condition: It is satisfied as the sample is less than $10\%$ of all people with Crohn’s disease.

Success/failure condition: It is satisfied because both are greater than ten as,

  $\begin{array}{l}np=335>10\\ n(1-p)=238>10\end{array}$

Thus, all the conditions are satisfied except the random condition. Therefore, now, to find the confidence interval we will use the calculator $\text{TI}-89$ . So, we have to first go to the $\text{Ints}$ menu and select $5:1-\text{PropZInt}$ . Then we have to enter the number of successes observed and the sample size. Then specify a confidence level. Then the interval will be calculated. Thus, we have,

  $\begin{array}{l}x=335\\ n=573\\ c=0.95=95\%\end{array}$

Thus, the confidence interval will be as:

  $(0.54429,0.62499)$ .

(b)

To determine

To interpret your interval in context.

Explanation of Solution

It is given in the question that medical journal reported patients suffering from Crohn’s disease responded positively to injections of the drug. Thus, the confidence interval will be as:

  $(0.54429,0.62499)$ .

Therefore, from the confidence interval calculated above we can conclude that we are $95\%$ confident that between $54.43\%\text{ and 62}\text{.5\%}$ of patients with Crohn’s disease will respond positively to the injections.

(c)

To determine

To explain carefully what “ $95\%$ confidence” means in this context.

Explanation of Solution

It is given in the question that medical journal reported patients suffering from Crohn’s disease responded positively to injections of the drug. Thus, the confidence interval will be as:

  $(0.54429,0.62499)$ .

Therefore, from the confidence interval calculated above we can conclude that we are $95\%$ confident that between $54.43\%\text{ and 62}\text{.5\%}$ of patients with Crohn’s disease will respond positively to the injections. So, “ $95\%$ confidence” means in this context that $95\%$ of interval created from samples will contain the true population proportions.