Chapter 9.2, Problem 1CYU

a.

To determine

State the null and alternate hypotheses.

Answer to Problem 1CYU

The hypotheses are given below:

Null hypothesis:

$H_0:p_1=p_2$

That is, there is no significant difference between the proportion of patients experienced relief from pain under drug-1 and proportion of patients experienced relief from pain under drug-2.

Alternate hypothesis:

$H_1:p_1>p_2$

That is, the proportion of patients experienced relief from pain under drug-1 is significantly greater than the proportion of patients experienced relief from pain under drug-2.

Explanation of Solution

It is given that among a sample of 100 patients under drug-1, 76 patients experienced relief from pain and among a sample of 200 patients under drug-2, 128 patients experienced relief from pain. The investigator wants to check whether the proportion of patients experienced relief from pain under drug-1 is significantly greater than the proportion of patients experienced relief from pain under drug-2. The level of significance is $\alpha =0.05$.

Hypothesis:

Hypothesis is an assumption about the parameter of the population, and the assumption may or may not be true.

Let $p_1$ be the population proportion of patients experienced relief from pain under drug-1 and $p_2$ be the population proportion of patients experienced relief from pain under drug-2.

Claim:

Here, the claim is whether the proportion of patients experienced relief from pain under drug-1 is significantly greater than the proportion of patients experienced relief from pain under drug-2.

The hypotheses are given below:

Null hypothesis:

Null hypothesis is a statement which is tested for statistical significance in the test. The decision criterion indicates whether the null hypothesis will be rejected or not in the favor of alternate hypothesis.

$H_0:p_1=p_2$

That is, there is no significant difference between the proportion of patients experienced relief from pain under drug-1 and proportion of patients experienced relief from pain under drug-2.

Alternate hypothesis:

Alternate hypothesis is contradictory statement of the null hypothesis

$H_1:p_1>p_2$

That is, the proportion of patients experienced relief from pain under drug-1 is significantly greater than the proportion of patients experienced relief from pain under drug-2.

b.

To determine

Find the proportion of patients experienced relief from pain under drug-1.

Find the proportion of patients experienced relief from pain under drug-2.

Answer to Problem 1CYU

The proportion of patients experienced relief from pain under drug-1 is $\underset{\_}{{\widehat{p}}_1=0.76}$.

The proportion of patients experienced relief from pain under drug-2 is $\underset{\_}{{\widehat{p}}_2=0.64}$

Explanation of Solution

Calculation:

Proportion of patients experienced relief from pain under drug-1:

The total number of patients under drug-1 is $n_1=100$ and the number of patients experienced relief from pain under drug-1 is $x_1=76$.

The proportion of patients experienced relief from pain under drug-1 is obtained as follows:

$\begin{array}{c}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{76}{100}\\ =0.76\end{array}$

Thus, the proportion of patients experienced relief from pain under drug-1 is 0.76.

Proportion of patients experienced relief from pain under drug-2:

The total number of patients under drug-2 is $n_2=200$ and the number of patients experienced relief from pain under drug-2 is $x_2=128$.

The proportion of patients experienced relief from pain under drug-2 is obtained as follows:

$\begin{array}{c}{\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{128}{200}\\ =0.64\end{array}$

Thus, the proportion of patients experienced relief from pain under drug-2 is 0.64.

c.

To determine

Find the value of t-test statistic.

Answer to Problem 1CYU

The value of test statistic is 2.100436.

Explanation of Solution

Calculation:

Estimate of pooled proportion:

The estimate of pooled proportion is obtained as follows:

$\begin{array}{c}\widehat{p}=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{76+128}{100+200}\\ =\frac{204}{300}\\ =0.68\end{array}$

Thus, the estimate of pooled proportion is 0.68.

From part (b), the sample proportions are ${\widehat{p}}_1=0.76$ and ${\widehat{p}}_2=0.64$.

Test statistic:

The test statistic for testing the difference between two proportions is,

$z=\frac{\left({\widehat{p}}_1-{\widehat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{\widehat{p}\left(1-\widehat{p}\right)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$

Under the null hypothesis, $\left(p_1-p_2\right)=0$.

The test statistic is obtained as follows,

$\begin{array}{c}z=\frac{\left({\widehat{p}}_1-{\widehat{p}}_2\right)-0}{\sqrt{\widehat{p}\left(1-\widehat{p}\right)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}\\ =\frac{0.76-0.64}{\sqrt{0.68\left(1-0.68\right)\left(\frac{1}{100}+\frac{1}{200}\right)}}\\ =\frac{0.12}{0.057131}\\ =2.100436\end{array}$

Thus, the test statistic is 2.100436.

d.

To determine

Find the P-value for the test statistic.

Answer to Problem 1CYU

The P-value for the test statistic is 0.01785.

Explanation of Solution

P-value:

Software procedure:

Step-by-step procedure to obtain the P-value using the MINITAB software:

• Choose Graph > Probability Distribution Plot.
• Choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X value and Right Tail for the region of the curve to shade.
• In X-value enter 2.100436.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the P-value is 0.01785.

Thus, the P-value is 0.01785.

e.

To determine

Interpret the P-value at the level of significance $\alpha =0.05$.

Answer to Problem 1CYU

There is enough evidence to reject the null hypothesis $H_0$.

Explanation of Solution

From part (d), the P-value is 0.01785.

Decision rule based on P-value:

If $P-\text{value}\le \alpha$, then reject the null hypothesis $H_0$.

If $P-\text{value}>\alpha$, then fail to reject the null hypothesis $H_0$.

Here, the level of significance is $\alpha =0.05$.

Conclusion based on P-value approach:

The P-value is 0.01785 and $\alpha$ value is 0.05.

Here, P-value is less than the $\alpha$ value.

That is, $0.01785\left(=P-\text{value}\right)<0.05\left(=\alpha \right)$.

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to reject the null hypothesis $H_0$.

f.

To determine

State the conclusion.

Answer to Problem 1CYU

There is enough evidence to conclude that the proportion of patients experienced relief from pain under drug-1 is significantly greater than the proportion of patients experienced relief from pain under drug-2.

Explanation of Solution

From part (e), it is known that the null hypothesis is rejected.

Thus, there is not enough evidence to conclude that the proportion of patients experienced relief from pain under drug-1 is significantly greater than the proportion of patients experienced relief from pain under drug-2.