Chapter 9, Problem 1PP

a)

To determine

To create: A contingency table for the provided data set. Also, draw the conclusion for the association between the variables.

Answer to Problem 1PP

It is impossible to comment on the association between the variables using the constructed contingency table

Explanation of Solution

Given information:

Number of men that do not drink coffee = 7890

Number of men who drink coffee more than 6 cups a day = 2492

Number of men who developed cancer in no coffee drinking group = 122

Number of men who developed cancer in high drinking coffee group = 19

Calculation:

Consider, the explanatory variable is Coffee Intake which is shown in the columns and the response variable is the cancer that is shown on the row.

So, the contingency table is constructed as:

Variables	Coffee Intake	No Coffee Intake	Total
Cancer	19	122	141
No Cancer	2473	7768	10237
Total	2492	7890	10382

Interpretation:

On the basis of above constructed contingency table, it is impossible to comment on the association between two variables.

b)

To determine

To find: The probability of advanced prostate cancer for the high-coffee group.

Answer to Problem 1PP

The probability is 0.00762.

Explanation of Solution

Calculation:

The probability is calculated as:

  $\begin{array}{c}P\left(\text{Cancer|Coffee Intake}\right)=\frac{P\left(\text{Cancer AND Coffee Intake}\right)}{P\left(\text{Coffee Intake}\right)}\\ =\frac{\frac{19}{10382}}{\frac{2492}{10382}}\\ =0.00762\end{array}$

The required probability is 0.00762.

c)

To determine

To find: The probability of advanced prostate cancer for the no-coffee group.

Answer to Problem 1PP

The probability is 0.01546.

Explanation of Solution

Calculation:

The probability is calculated as:

  $\begin{array}{c}P\left(\text{Cancer|No coffee Intake}\right)=\frac{P\left(\text{Cancer AND No Coffee Intake}\right)}{P\left(\text{Coffee Intake}\right)}\\ =\frac{\frac{122}{10382}}{\frac{7890}{10382}}\\ =0.01546\end{array}$

The required probability is 0.01546.

d)

To determine

To find: The relative risk of advanced prostate cancer, comparing the treatment and control groups

Answer to Problem 1PP

The probability is 0.01546.

Explanation of Solution

Calculation:

Consider the following table to calculate the relative Risk:

	Coffee	No Coffee
Cancer	19 (a)	122 (b)
No Cancer	2473 (c)	7768 (d)

The relative risk is computed as:

  $\begin{array}{c}\text{RelativeRisk}=\frac{\frac{a}{a+c}}{\frac{b}{b+d}}\\ =\frac{\frac{19}{2492}}{\frac{122}{7890}}\\ =0.493\\ \simeq 0.50\end{array}$

Hence, the required relative risk is 0.50.

e)

To determine

To find: The odds of advanced prostate cancer for high-coffee group.

Answer to Problem 1PP

The odds are 0.0077.

Explanation of Solution

Calculation:

The odds could be calculated as:

  $\begin{array}{c}\text{Odds}=\frac{a}{c}\\ =\frac{19}{2473}\\ =0.00768\\ \simeq 0.0077\end{array}$

The required odds are 0.0077.

f)

To determine

To find: The odds of advanced prostate cancer for non-coffee group.

Answer to Problem 1PP

The odds are 0.0157.

Explanation of Solution

Calculation:

The odds could be calculated as:

  $\begin{array}{c}\text{Odds}=\frac{b}{d}\\ =\frac{122}{7764}\\ =0.0157\end{array}$

The required odds are 0.0157.

g)

To determine

To compare: The odds ratio of advanced prostate cancer comparing two groups.

Explanation of Solution

Calculation:

The odds ratio is cancer is defined as ratio of the odds of cancer in the coffee- drinking group and the odds of cancer in the non-coffee drinking group.

The odds ratio of cancer for comparing both the groups is:

  $\begin{array}{c}\text{Odds ratio}=\frac{\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}{\raisebox{1ex}{$b$}\!\left/ \!\raisebox{-1ex}{$d$}\right.}\\ =\frac{ad}{bc}\\ =\frac{19\times 7764}{122\times 2473}\\ =0.4889\end{array}$

The required odds ratio is 0.4889.

h)

To determine

To find: The log odds ratio comparing these two groups.

Answer to Problem 1PP

The log odds ratio is -0.7156.

Explanation of Solution

Calculation:

The log-odds ratio of cancer for comparing both the groups is:

  $\begin{array}{c}\text{Log Odds ratio}=\ln (OddsRatio)\\ =\ln (0.4889)\\ =-0.7156\end{array}$

The required log odds ratio is -0.7156.

i)

To determine

To find: The standard error of log odds ratio in the provided case.

Answer to Problem 1PP

The standard error is 0.2477.

Explanation of Solution

Calculation:

Standard Error of the log-odds ratio is

  $\begin{array}{c}\text{Standard error}=\sqrt{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}\\ =\sqrt{\frac{1}{19}+\frac{1}{122}+\frac{1}{2473}+\frac{1}{7764}}\\ =\sqrt{0.06134}\\ =0.2477\end{array}$

The required standard error is 0.2477.

j)

To determine

To find: The 95% confidence interval for the log odds ratio.

Answer to Problem 1PP

The confidence interval is (-1.2011, -0.2301)

Explanation of Solution

Calculation:

The critical value for 95% confidence level is 1.96.

The confidence interval for the log odds ratio is calculated as:

  $\begin{array}{c}CI=\text{Log odds ratio}\pm \text{z}\times \text{SE}\\ \text{=}\left(-0.7156\right)\pm 1.96\left(0.2477\right)\\ =\left(-1.2011,-0.2301\right)\end{array}$

The required confidence interval is (-1.2011, -0.2301)

k)

To determine

To find: The 95% confidence interval for the odds ratio.

Answer to Problem 1PP

The confidence interval is (0.3,0.79)

Explanation of Solution

Calculation:

The confidence interval for odds ratio is exponential of confidence interval of log-odds ratio.

The confidence interval is calculated as:

  $\begin{array}{c}CI=\text{Exponential}\left(\text{Log odds ratio}\pm \text{z}\times \text{SE}\right)\\ \text{=Exponential}\left(\left(-0.7156\right)\pm 1.96\left(0.2477\right)\right)\\ =\left(0.3,0.79\right)\end{array}$

The required confidence interval is (0.3,0.79).

l)

To determine

To interpret: The computed confidence interval for the odd ratio. Check if it is consistent with the case that coffee drinking and developing cancer are independent. Also, check if coffee consumption is related to the increased or decreased probability of advanced prostate cancer.

Explanation of Solution

Calculation:

The estimated odds ratio lies in the $95\%$ confidence interval of $(0.3,0.79)$ . But the true Odds Ratio, 1, does not lie in the estimated interval. So, the cancer rates in the Coffee drinking group and the non-coffee drinking group is not identical.

The probability of cancer in coffee drinking group is

  $\begin{array}{c}P\left(\text{No Cancer}\cap \text{Coffee Intake}\right)=\frac{19}{2492}\\ =0.0076\end{array}$

The probability of not developing cancer in coffee drinking group is:

  $\begin{array}{c}P\left(\text{No Cancer}\cap \text{Coffee Intake}\right)=\frac{2473}{2492}\\ =0.9924\end{array}$

From above calculation, it is clear that there is significant decrease in developing cancer in coffee drinking group.

Conclusion:

Yes, the coffee consumption is related to the increased or decreased probability of advanced prostate cancer.