Chapter 5.5, Problem 3E

Angioplasty is a medical procedure in which an obstructed blood vessel is widened. In some cases, a wire mesh tube, called a stent, is placed in the vessel to help it remain open. The article “Long-term Outcomes of Patients Receiving Drug-eluting Stents” (A. Philpott, D. Southern, et al., Canadian Medical Association Journal, 2009:167–174) presents the results of a study comparing the effectiveness of a bare metal stent with one that that has been coated with a drug designed to prevent reblocking of the vessel. A total of 5320 patients received bare metal stents, and of these, 841 needed treatment for reblocking within a year. A total of 1120 received drug coated stents, and 134 of them required treatment within a year.

1. a. Find a 98% confidence interval for the differences between the proportions for drug coated stents and bare metal stents.
2. b. Suppose that additional patients are to be treated in order to increase the precision of the confidence interval. Three sampling plans are being considered. In the first plan, 1000 additional patients will be treated with bare metal stents. In the second plan, 500 additional patients will be treated with drug coated stents. In the third plan, 500 additional patients will be treated with bare metal stents and 250 additional patients will be treated with drug coated stents. Which plan is most likely to provide the greatest increase in the precision of the confidence interval? Explain.

To determine

Find the 98% confidence interval for the difference between the proportions for drug coated stents and bare metal stents.

Answer to Problem 3E

The 98% confidence interval for the difference between the proportions for drug coated stents and bare metal stents is $\underset{\_}{\left(0.0125,0.0633\right)}$.

Explanation of Solution

Given info:

$X=841$, $n_X=5,320$, $Y=134$, $n_Y=1,120$

Calculation:

Proportion:

The value of proportion is,

$\begin{array}{c}{\tilde{p}}_X=\frac{X+1}{n_X+2}\\ =\frac{841+1}{5,320+2}\\ =0.1582\end{array}$

$\begin{array}{c}{\tilde{p}}_Y=\frac{Y+1}{n_Y+2}\\ =\frac{134+1}{1,120+2}\\ =0.1203\end{array}$

Confidence interval:

Step-by-step procedure to obtain the confidence interval using the MINITAB software:

• Choose Stat > Basic Statistics > 2 Proportion.
• Choose Summarized data.
• In First sample, enter Number of events, enter 842 and enter Number of trials, enter 5322.
• In Second sample, enter Number of events, enter 135 and enter Number of trials, enter 1,122.
• Check Options; enter Confidence level as 98%.
• Choose not equal in alternative.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the 98% confidence interval for the difference between the proportions for drug coated stents and bare metal stents is $\underset{\_}{\left(0.0125,0.0633\right)}$.

To determine

Identify which plan is most likely to provide the greatest increase in the precision of the confidence interval.

Answer to Problem 3E

The second plan 500 additional patients will be treated with drug coated stents is most likely to provide the greatest increase in the precision of the confidence interval.

Explanation of Solution

Calculation:

If the 1,000 additional patients will be treated with bare metal stents, then $n_X=6,320$, and $n_Y=1,120$.

The standard deviation for the difference between the proportions is,

$\begin{array}{c}\sqrt{\frac{p_X\left(1-p_X\right)}{n_X+2}+\frac{p_Y\left(1-p_Y\right)}{n_Y+2}}=\sqrt{\frac{0.1582\left(1-0.1582\right)}{6,320+2}+\frac{0.1203\left(1-0.1203\right)}{1,120+2}}\\ =0.0107\end{array}$

The width of 98% confidence interval is,

$\pm 2.33\left(0.0107\right)=\pm \text{0}\text{.0249}$

If the 500 additional patients will be treated with drug coated stents, then $n_X=5,320$, and $n_Y=1,620$.

The standard deviation for the difference between the proportions is,

$\begin{array}{c}\sqrt{\frac{p_X\left(1-p_X\right)}{n_X+2}+\frac{p_Y\left(1-p_Y\right)}{n_Y+2}}=\sqrt{\frac{0.1582\left(1-0.1582\right)}{5,320+2}+\frac{0.1203\left(1-0.1203\right)}{1,620+2}}\\ =0.0097\end{array}$

The width of 98% confidence interval is,

$\pm 2.33\left(0.0097\right)=\pm \text{0}\text{.0226}$

If the 500 additional patients will be treated with bare metal stents and 250 additional patients will be treated with drug coated stents, then $n_X=5,820$, and $n_Y=1,370$.

The standard deviation for the difference between the proportions is,

$\begin{array}{c}\sqrt{\frac{p_X\left(1-p_X\right)}{n_X+2}+\frac{p_Y\left(1-p_Y\right)}{n_Y+2}}=\sqrt{\frac{0.1582\left(1-0.1582\right)}{5,820+2}+\frac{0.1203\left(1-0.1203\right)}{1,370+2}}\\ =0.0100\end{array}$

The width of 98% confidence interval is,

$\pm 2.33\left(0.0100\right)=\pm \text{0}\text{.0233}$

From the result, it can be observed that the width of 98% confidence interval for second plan is lesser when compared to other two plans. Hence, the second plan 500 additional patients will be treated with drug coated stents is most likely to provide the greatest increase in the precision of the confidence interval.