Chapter 10, Problem 57CE

a.

To determine

State the hypotheses for a one-tailed test.

Explanation of Solution

The given information is that the medical center implanted defibrillators in 742 patients after a heart attack, 104 patients with defibrillators had diet. The total of 490 patients without the implant, 98 patients without defibrillators. That is, $x_1=104$, $n_1=742$, $x_2=98$ and $n_2=490$.

Let ${\pi }_1$ be the proportion of died with the defibrillator and ${\pi }_2$ be the proportion of died without the defibrillator.

Here, the direction of the test is left-tailed.

State the hypotheses:

Null hypothesis:

$H_0:{\pi }_1-{\pi }_2\ge 0$

That is, the proportion of died without the defibrillator implanted is not higher than the proportion of died with the defibrillator.

Alternative hypothesis:

$H_1:{\pi }_1-{\pi }_2<0$

That is, the proportion of died without the defibrillator implanted is higher than the proportion of died with the defibrillator.

b.

To determine

Find the test statistic and p-value.

Answer to Problem 57CE

The test statistic is –2.78.

The p-value is 0.0027.

Explanation of Solution

Calculation:

Sample proportions:

First sample:

$\begin{array}{c}{p}_1=\frac{x_1}{n_1}\\ =\frac{104}{742}\\ =0.14\end{array}$

Second sample:

$\begin{array}{c}{p}_2=\frac{x_2}{n_2}\\ =\frac{98}{490}\\ =0.2\end{array}$

Pooled proportion:

$\begin{array}{c}{p}_c=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{104+98}{742+490}\\ =\frac{202}{1,232}\\ =0.164\end{array}$

Thus, the pooled proportion is 0.164.

Test statistic:

$\begin{array}{c}{z}_{\text{calc}}=\frac{p_1-p_2}{\sqrt{p_c\left(1-p_c\right)\left[\frac{1}{n_1}+\frac{1}{n_2}\right]}}\\ =\frac{0.14-0.2}{\sqrt{0.164\left(1-0.164\right)\left[\frac{1}{742}+\frac{1}{490}\right]}}\\ =\frac{-0.06}{\sqrt{0.164\left(0.836\right)\left[0.00135+0.00204\right]}}\\ =\frac{-0.06}{\sqrt{0.1371\left(0.00339\right)}}\\ =-2.78\end{array}$

Thus, the test statistic is –2.78.

p-value:

Software procedure:

Step-by-step software procedure to obtain p-value using EXCEL is as follows:

• • Open an EXCEL file.
• • In cell A1, enter the formula “=NORM.S.DIST (–2.78,1)”.
• Output using EXCEL software is given below:

From the output, the p-value is 0.0027.

Decision rule:

If $p\text{-value}<\alpha$, then reject the null hypothesis.

Conclusion for p-value method:

Here, the p-value is less than the level of significance.

That is, $p\text{-value}\left(=0.0027\right)<\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected.

Thus, there is evidence to infer that the proportion of died without the defibrillator implanted is higher than the proportion of died with the defibrillator.

c.

To determine

Check whether the normality of $p_1-p_2$ can be assumed or not.

Answer to Problem 57CE

The normality of $p_1-p_2$ can be assumed.

Explanation of Solution

Calculation:

Rule for normality:

• • Rule 1: $n_1{p}_1\ge 10$
• • Rule 2: $n_1\left(1-p_1\right)\ge 10$
• • Rule 3: $n_2{p}_2\ge 10$
• • Rule 4: $n_2\left(1-p_2\right)\ge 10$

Check the rule:

Rule 1: $n_1{p}_1\ge 10$

$\begin{array}{c}{n}_1{p}_1=742\left(0.14\right)\\ =103.88\left(>10\right)\end{array}$

Rule 2: $n_1\left(1-p_1\right)\ge 10$

$\begin{array}{c}{n}_1\left(1-p_1\right)=742\left(1-0.14\right)\\ =638.12\left(>10\right)\end{array}$

Rule 3: $n_2{p}_2\ge 10$

$\begin{array}{c}{n}_2{p}_2=490\left(0.2\right)\\ =98\left(>10\right)\end{array}$

Rule 4: $n_2\left(1-p_2\right)\ge 10$

$\begin{array}{c}{n}_2\left(1-p_2\right)=490\left(1-0.2\right)\\ =392\left(>10\right)\end{array}$

Since $n_1{p}_1$, $n_1\left(1-p_1\right)$, $n_2{p}_2$ and $n_2\left(1-p_2\right)$ are greater than 10, the normality of the sample proportion can be assumed.

d.

To determine

Explain why such devices might not be widely implanted in heart attack patients.

Explanation of Solution

Some of devices “exercise habits, smoking habits, nutrition, weight and so on” might not be widely implanted in heart attack patients because these devices might be effects the decision.