Chapter 6.6, Problem 70E

a.

To determine

Construct a table that represents the frequency of respondent in each of the six cells.

Answer to Problem 70E

The table of frequency of respondent is given by:

	Use alternative therapies	Does not use alternative therapies	Total
High School or less	315	7,005	7,320
College-1 to 4 years	393	4,400	4,793
College-5 or more year	120	975	1,095
Total	828	12,380	13,208

Explanation of Solution

Calculation:

The given information is a percentage distribution of education levels. About 7,320 people have education level high school or less, 4,793 people with 1 to 4 years of college, and 1,095 people with five or more years of college.

It known that 4.3% of the people using alternative therapies and have education level high school or less.

The frequency of people using alternative therapies and have education level high school or less is given by:

$\begin{array}{c}N\left(\begin{array}{l}\text{Alternative therapy and education}\\ \text{ level is high school or less}\end{array}\right)=4.3\%\text{ of 7,320}\\ \text{=}\frac{4.3}{100}\times 7,320\\ =314.76\\ =315\end{array}$

Thus, the 315 people use alternative therapies and have education level high school or less.

Similarly, the other frequencies are obtained as shown in the table:

	Use alternative therapies	Does not use alternative therapies	Total
High School or less	315	$7,320-315=7,005$	7,320
College-1 to 4 years	$\left(0.082\right)\times \left(4,793\right)=393$	$4,793-393=4,400$	4,793
College-5 or more year	$\left(0.11\right)\times \left(1,095\right)=120$	$1,095-120=975$	1,095
Total	828	12,380	13,208

b.

To determine

Construct a table that probabilities of respondent in each of the six cells.

Answer to Problem 70E

The table of respondent is given by:

	Use alternative therapies	Does not use alternative therapies	Total
High School or less	0.024	0.530	0.554
College-1 to 4 years	0.030	0.333	0.363
College-5 or more year	0.009	0.074	0.083
Total	0.063	0.937	1

Explanation of Solution

Calculation:

The probability of the event T can be obtained by the formula:

$P\left(T\right)=\frac{\text{Number of outcomes in }T}{\text{Number of outcomes in}\text{sample space}}$

Therefore, the probability of people using alternative therapies and have education level high school or less is given by:

$\begin{array}{c}P\left(\begin{array}{l}\text{Alternative therapy and education}\\ \text{ level is high school or less}\end{array}\right)=\frac{315}{13,208}\\ =0.024\end{array}$

Thus, the probability that people have education level high school or less and use alternative therapies is 0.024.

Similarly, the other probabilities are obtained as shown in the table:

	Use alternative therapies	Does not use alternative therapies	Total
High School or less	0.024	$\frac{7,005}{13,208}=0.530$	$\frac{7,320}{13,208}=0.554$
College-1 to 4 years	$\frac{393}{13,208}=0.030$	$\frac{4,400}{13,208}=0.333$	$\frac{4,793}{13,208}=0.363$
College-5 or more year	$\frac{120}{13,208}=0.009$	$\frac{975}{13,208}=0.074$	$\frac{1,095}{13,208}=0.083$
Total	$\frac{828}{13,208}=0.063$	$\frac{12,380}{13,208}=0.937$	1.000

c.

To determine

1. i.) Compute the probability that selected individual has 5 or more years of college.
2. ii.) Calculate the probability that selected individual uses alternative therapies.
3. iii.) Calculate the probability that selected individual uses alternative therapies given that the person has 5 or more years of college.
4. iv.) Calculate the probability that selected individual uses alternative therapies given that the person has an educational level of high school or less.
5. v.) Calculate the probability that selected individual uses alternative therapies has an educational level of high school or less.
6. i.) Calculate the probability that randomly selected individual with some college use alternative therapies.

Answer to Problem 70E

1. i.) The probability that selected individual has 5 or more years of college is 0.083.
2. ii.) The probability that selected individual uses alternative therapies is 0.063.
3. i.) The probability that selected individual uses alternative therapies given that the person has 5 or more years of college is 0.110.
4. ii.) The probability that selected individual uses alternative therapies given that the person has an educational level of high school or less is 0.043.
5. iii.) The probability that selected individual uses alternative therapies has an educational level of high school or less is 0.380.
6. iv.) The probability that randomly selected individual with some college use alternative therapies is 0.087.

Explanation of Solution

Calculation:

(i.) and (ii):

From the table of probabilities, it is clear that, the probability that selected individual has 5 or more years of college is 0.083 and that of selected individual uses alternative therapies is 0.063.

iii.

The required probability is given as follows:

$P\left(\text{alternative therapy| 5 or more years of college}\right)=\frac{P\left(\begin{array}{l}\text{alternative therapy}\text{and}\text{5 or }\\ \text{more years of college }\end{array}\right)}{P\left(\text{5 or more years of college}\right)}$

$P\left(\begin{array}{l}\text{alternative therapy}\text{and}\text{5 or }\\ \text{more years of college }\end{array}\right)=0.009\text{and }P\left(\text{5 or more years of college }\right)=0.083$.

The required probability is given as follows:

$\begin{array}{c}P\left(\text{alternative therapy| 5 or more years of college}\right)=\frac{0.009}{0.083}\\ =0.108\\ \approx 0.110\end{array}$

Thus, the probability that selected individual uses alternative therapies given that the person has 5 or more years of college is 0.110.

iv.

The required probability is given as follows:

$P\left(\text{alternative therapy| high school or less}\right)=\frac{P\left(\text{alternative therapy}\text{and}\text{high school or less}\right)}{P\left(\text{high school or less}\right)}$

$P\left(\begin{array}{l}\text{alternative therapy}\text{and}\\ \text{high school or less}\end{array}\right)=0.024\text{and }P\left(\text{high school or less}\right)=0.554$.

The required probability is given as follows:

$\begin{array}{c}P\left(\text{alternative therapy| high school or less}\right)=\frac{0.024}{0.554}\\ =0.043\end{array}$

Thus, the probability that selected individual uses alternative therapies given that the person has an educational level of high school or less is 0.043.

v.

The required probability is given as follows:

$P\left(\text{high school or less| alternative therapy}\right)=\frac{P\left(\text{alternative therapy}\text{and}\text{high school or less}\right)}{P\left(\text{alternative therapy}\right)}$

$P\left(\begin{array}{l}\text{alternative therapy}\text{and}\\ \text{high school or less}\end{array}\right)=0.024\text{and }P\left(\text{alternative therapy}\right)=0.063$.

The required probability is given as follows:

$\begin{array}{c}P\left(\text{high school or less| alternative therapy}\right)=\frac{0.024}{0.063}\\ =0.380\end{array}$

Thus, the probability that selected individual uses alternative therapies has an educational level of high school or less is 0.380.

vi.

The required probability is given as follows:

$P\left(\text{alternative therapy| college}\right)=\frac{P\left(\text{alternative therapy}\text{and}\text{college}\right)}{P\left(\text{college}\right)}$

$\begin{array}{c}P\left(\text{alternative therapy}\text{and}\text{college}\right)=0.030+0.009\\ =0.039\\ P\left(\text{college}\right)=0.363+0.083\\ =0.446\end{array}$.

The required probability is given as follows:

$\begin{array}{c}P\left(\text{alternative therapy| college}\right)=\frac{0.039}{0.446}\\ =0.087\end{array}$

Thus, the probability that randomly selected individual with some college use alternative therapies is 0.087.

d.

To determine

Check whether the events H and A are independent or not.

Answer to Problem 70E

Events H and A are not independent.

Explanation of Solution

Calculation:

Event H denotes that selected individual has an education level of high school or less and A denotes that selected individual uses alternative therapies.

Independent Events:

Two events H and A independent if $P\left(A|H\right)=P\left(A\right)$.

It was already found that that, $P\left(A|H\right)=0.043$ and $P\left(A\right)=0.063$.

Here,

$\begin{array}{c}P\left(A|H\right)=0.043\\ \ne P\left(A\right)\end{array}$

Therefore, the events H and A are not independent.