Chapter 5, Problem 94CE

Refer to the contingency table shown below. (a) Calculate each probability (i–vi) and explain in words what it means. (b) Do you see evidence that smoking and race are not independent? Explain. (c) Do the smoking rates shown here correspond to your experience? (d) Why might public health officials be interested in this type of data?

1. i. P(S)
2. ii. P(W)
3. iii. P(S|W)
4. iv. P(S|B)
5. v. P(S and W)
6. vi. P(N and B)

Smoking by Race for Males Aged 18–24

To determine

Calculate each probability (i-vi) and explain it meaning in words.

Answer to Problem 94CE

i. The probability $P\left(S\right)$ is 0.32 and the meaning is the probability of a male 18-24 smoking is 0.32.

ii. The probability $P\left(W\right)$ is 0.85 and the meaning is the probability of a male 18-24 is white is 0.85.

iii. The probability $P\left(S|W\right)$ is 0.3412 and the meaning is the probability of white male is smoker is 0.3412.

iv. The probability $P\left(S|B\right)$ is 0.20 and the meaning is the probability of black male is smoker is 0.20.

v. The probability $P\left(S\text{ and W}\right)$ is 0.290 and the meaning is the probability of a male 18-24 is smoker and white is 0.290.

vi. The probability $P\left(N\text{ and }B\right)$ is 0.12 and the meaning is the probability of a male 18-24 is not smoking and white is 0.12.

Explanation of Solution

Calculation:

The given table shows that the smoking by race for males aged 18-24.

The given contingency table is,

	Smoker (S)	Nonsmoker (N)	Row Total
White (W)	290	560	850
Black (B)	30	120	150
Column Total	320	680	1,000

For (i)$P\left(S\right)$:

The formula for finding the probability $P\left(S\right)$ is,

$P\left(S\right)=\frac{\text{Frequency for the class }S}{\text{Total frequencies in the distribution}}$

Substitute 320 for ‘Frequency for the class S’ and 1,000 for ‘Total frequencies in the distribution’

$\begin{array}{c}P\left(S\right)=\frac{320}{1,000}\\ =0.320\end{array}$

Therefore, the probability $P\left(S\right)$ is 0.32 and the meaning is the probability of a male 18-24 smoking is 0.32.

For (ii) $P\left(W\right)$:

The formula for finding the probability $P\left(W\right)$ is,

$P\left(W\right)=\frac{\text{Frequency for the class }W}{\text{Total frequencies in the distribution}}$

Substitute 850 for ‘Frequency for the class W’ and 1,000 for ‘Total frequencies in the distribution’

$\begin{array}{c}P\left(W\right)=\frac{850}{1,000}\\ =0.85\end{array}$

Therefore, the probability $P\left(W\right)$ is 0.85 and the meaning is the probability of a male 18-24 is white is 0.85.

For (iii) $P\left(S|W\right)$:

The formula for finding the probability $P\left(S|W\right)$ is,

$P\left(S|W\right)=\frac{\text{Frequency for the class }S\text{ and }W}{\text{Frequency for the class }W}$

Substitute 290 for ‘Frequency for the class S and W’ and 850 for ‘Frequency for the class W’,

$\begin{array}{c}P\left(S|W\right)=\frac{290}{850}\\ =0.3412\end{array}$

Therefore, the probability $P\left(S|W\right)$ is 0.3412 and the meaning is the probability of white male is smoker is 0.3412.

For (iv) $P\left(S|B\right)$:

The formula for finding the probability $P\left(S|B\right)$ is,

$P\left(S|B\right)=\frac{\text{Frequency for the class }S\text{ and }B}{\text{Frequency for the class }B}$

Substitute 30 for ‘Frequency for the class S and B’ and 150 for ‘Frequency for the class B’,

$\begin{array}{c}P\left(S|B\right)=\frac{30}{150}\\ =0.20\end{array}$

Therefore, the probability $P\left(S|B\right)$ is 0.20 and the meaning is the probability of black male is smoker is 0.20.

For (v) $P\left(S\text{ and W}\right)$:

The formula for finding the probability $P\left(S\text{ and W}\right)$ is,

$P\left(S\text{ and W}\right)=\frac{\text{Frequency for the class }S\text{ and }W}{\text{Total frequencies in the distribution}}$

Substitute 290 for ‘Frequency for the class S and W’ and 1,000 for ‘Total frequencies in the distribution’,

$\begin{array}{c}P\left(S\text{ and W}\right)=\frac{290}{1,000}\\ =0.290\end{array}$

Therefore, the probability $P\left(S\text{ and W}\right)$ is 0.290 and the meaning is the probability of a male 18-24 is smoker and white is 0.290.

For (vi) $P\left(N\text{ and }B\right)$:

The formula for finding the probability $P\left(N\text{ and }B\right)$ is,

$P\left(N\text{ and }B\right)=\frac{\text{Frequency for the class }N\text{ and }B}{\text{Total frequencies in the distribution}}$

Substitute 120 for ‘Frequency for the class N and B’ and 1,000 for ‘Total frequencies in the distribution’,

$\begin{array}{c}P\left(N\text{ and }B\right)=\frac{120}{1,000}\\ =0.12\end{array}$

Therefore, the probability $P\left(N\text{ and }B\right)$ is 0.12 and the meaning is the probability of a male 18-24 is not smoking and white is 0.12.

To determine

Check whether there is evidence that smoking and race are not independent or not. Explain the reason.

Answer to Problem 94CE

Yes, there is evidence that smoking and race are not independent because $P\left(S\text{ and }W\right)\ne P\left(S\right)P\left(W\right)$ and $P\left(S\text{ and }B\right)\ne P\left(S\right)P\left(B\right)$.

Explanation of Solution

Calculation:

Special law of multiplication:

If two events A and B are independent, then

$P\left(A\cap B\right)=P\left(A\right)P\left(B\right)$

Consider smoking type smoker and race type as white.

From part (a), $P\left(S\text{ and }W\right)=0.290$.

The formula for checking independence is,

$\begin{array}{c}P\left(S\text{ and }W\right)=P\left(S\right)P\left(W\right)\\ 0.290=\left(\frac{\text{Frequency for the class }S}{\text{Total frequencies in the distribution}}\right)\left(\frac{\text{Frequency for the class W}}{\text{Total frequencies in the distribution}}\right)\\ =\frac{320}{1,000}\times \frac{850}{1,000}\\ =\frac{272,000}{1,000,000}\\ 0.290\ne 0.272\end{array}$

Consider smoking type smoker and race type as black.

The formula for checking independence is,

$\begin{array}{c}P\left(S\text{ and }B\right)=P\left(S\right)P\left(B\right)\\ \left(\frac{\text{Frequency for the class }S\text{ and }B}{\text{Total frequencies in the distribution}}\right)=\left(\begin{array}{l}\left(\frac{\text{Frequency for the class }S}{\text{Total frequencies in the distribution}}\right)\times \\ \left(\frac{\text{Frequency for the class }B}{\text{Total frequencies in the distribution}}\right)\end{array}\right)\\ \frac{30}{1,000}=\frac{320}{1,000}\times \frac{150}{1,000}\\ 0.30=\frac{48,000}{1,000,000}\\ 0.30\ne 0.048\end{array}$

Here, it is observed that $P\left(S\text{ and }W\right)\ne P\left(S\right)P\left(W\right)$ and $P\left(S\text{ and }B\right)\ne P\left(S\right)P\left(B\right)$. Therefore, the there is evidence that smoking and race are not independent.

To determine

Check whether the given smoking rates shown here correspond to the person experience.

Answer to Problem 94CE

Yes, the given smoking rates shown here correspond to the person experience.

Explanation of Solution

Answers may vary: one of the answers is given below.

In experience, it is found that smoking has great influence on the racing.

Also in part (b), it is observed that the smoking and race are dependent. That is smoking is influencing the race. Thus, both experience and the smoking rates tells that same.

Therefore, the given smoking rates shown here correspond to the person experience.

To determine

Explain the reason for public health officials are interested in the given type of data.

Explanation of Solution

From part (b), it can be observed that the smoking is dependent on race.

Here, the public health officials will design the special programs for the given types of data when the smoking is dependent on race.