Chapter 6.6, Problem 8E

To determine

Check whether there is evidence to conclude that the percentage of people who have had colonoscopies is greater in those without colorectal cancer.

Answer to Problem 8E

There is evidence to conclude that the percentage of people who have had colonoscopies is greater in those without colorectal cancer.

Explanation of Solution

Given info:

In a sample of 10,292 people diagnosed with colorectal cancer, 7.0% had previously had a colonoscopy, and in a sample of 51,460 people without colorectal cancer, 9.8% had previously had a colonoscopy. That is, ${\widehat{p}}_X=0.070$, $n_X=\text{1}0,\text{292 }$, ${\widehat{p}}_Y=0.098$ and $n_Y=\text{51},\text{46}0$.

Calculation:

Assumptions:

• The samples are independent.
• There are at least 10 success and 10 failures for each samples.

Here, the samples are independent.

Verify the condition:

With colorectal cancer:

Condition 1: $n_X{p}_X\ge 10$

$\begin{array}{c}{n}_X{p}_X=\text{1}0,\text{292 }\left(0.070\right)\\ =\text{720}\text{.44}\left(>10\right)\end{array}$

Condition 2: $n_X\left(1-p_X\right)\ge 10$

$\begin{array}{c}{n}_X\left(1-p_X\right)=\text{1}0,\text{292 }\left(1-0.070\right)\\ =\text{9,571}\text{.56}\left(>10\right)\end{array}$

Without colorectal cancer:

Condition 1: $n_Y{p}_Y\ge 10$

$\begin{array}{c}{n}_Y{p}_Y=\text{51},\text{46}0\left(0.098\right)\\ =\text{5,043}\text{.08}\left(>10\right)\end{array}$

Condition 2: $n_Y\left(1-p_Y\right)\ge 10$

$\begin{array}{c}{n}_Y\left(1-p_Y\right)=\text{51},\text{46}0\left(1-0.098\right)\\ =\text{46,416}\text{.92}\left(>10\right)\end{array}$

Thus, the conditions are satisfied.

The pooled proportion is,

$\begin{array}{c}\widehat{p}=\frac{n_X{\widehat{p}}_X+n_Y{\widehat{p}}_Y}{n_X+n_Y}\\ =\frac{\left(10,292\times 0.070\right)+\left(51,460\times 0.098\right)}{10,292+51,460}\\ =\frac{\text{720}\text{.44}+\text{5,043}\text{.08}}{\text{61,752}}\\ =\text{0}\text{.0933333}\end{array}$

State the test hypotheses.

Null hypotheses:

$H_0:p_{\text{X}}-p_{\text{Y}}\ge 0$

Alternative hypothesis:

$H_1:p_{\text{X}}-p_{\text{Y}}<0$

Test statistic:

$\begin{array}{c}z=\frac{{\widehat{p}}_X-{\widehat{p}}_Y}{\sqrt{\widehat{p}\left(1-\widehat{p}\right)\left(\frac{1}{n_X}+\frac{1}{n_Y}\right)}}\\ =\frac{0.070-0.098}{\sqrt{0.0933333\left(1-0.0933333\right)\left(\frac{1}{10,292}+\frac{1}{51,460}\right)}}\\ =\frac{-\text{0}\text{.028}}{\text{0}\text{.0031411}}\\ =-8.91\end{array}$

Thus, the test statistic is –8.91.

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 Left Tail for the region of the curve to shade.
• Enter the data value as –8.91.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the P-value is approximately 0.000.

Decision rule:

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

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

Conclusion:

Here, the P-value is less than the level of significance, 0.05.

That is, $P\text{-value}\left(=0.000\right)<\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected.

Thus, there is evidence to conclude that the percentage of people who have had colonoscopies is greater in those without colorectal cancer.