Chapter PV, Problem 12RE

(a)

To determine

To explain do these data meet the assumptions necessary for inference and what would you like to know that you don’t.

Explanation of Solution

In the question it is given that in July the American journal of clinical nutrition reported that $42\%\text{ of 1546}$ African-American women studied had vitamin D deficiency. Thus, let us check the conditions and assumptions necessary for inference as:

Random condition: It is not satisfied as it is not given that the sample was randomly selected from the population.

Independent condition: It is satisfied as we can assume that the sample is independent.

  $10\%$ condition: It is satisfied as $1546$ African-American women are less than $10\%$ of all African-American women.

Success/failure condition: It is satisfied because both are greater than ten as,

  $\begin{array}{l}np=649.32>10\\ n(1-p)=896.68>10\end{array}$

Thus, all the conditions are satisfied except the random condition. Thus, we would like to know whether the sample was randomly selected from the population.

(b)

To determine

To create a $95\%$ confidence interval.

Answer to Problem 12RE

  $(0.39519,0.44439)$ .

Explanation of Solution

In the question it is given that in July the American journal of clinical nutrition reported that $42\%\text{ of 1546}$ African-American women studied had vitamin D deficiency. Therefore, now, to find the confidence interval we will use the calculator $\text{TI}-89$ . So, we have to first go to the $\text{Ints}$ menu and select $5:1-\text{PropZInt}$ . Then we have to enter the number of successes observed and the sample size. Then specify a confidence level. Then the interval will be calculated. Thus, we have,

  $\begin{array}{l}x=649.32\\ n=1546\\ c=0.95=95\%\end{array}$

Thus, the confidence interval will be as:

  $(0.39519,0.44439)$ .

(c)

To determine

To interpret the interval in this context.

Explanation of Solution

In the question it is given that in July the American journal of clinical nutrition reported that $42\%\text{ of 1546}$ African-American women studied had vitamin D deficiency. Thus, the confidence interval will be as:

  $(0.39519,0.44439)$ .

Thus, we can conclude about the confidence interval that we are $95\%$ confident that the true proportion of African-American women with Vitamin D deficiency is between $39.5\%\text{ and 44}\text{.4\%}$ .

(d)

To determine

To explain in this context what “ $95\%$ confidence” means.

Explanation of Solution

In the question it is given that in July the American journal of clinical nutrition reported that $42\%\text{ of 1546}$ African-American women studied had vitamin D deficiency. Thus, the confidence interval will be as:

  $(0.39519,0.44439)$ .

Thus, we can conclude about the confidence interval that we are $95\%$ confident that the true proportion of African-American women with Vitamin D deficiency is between $39.5\%\text{ and 44}\text{.4\%}$ . Thus, in this context “ $95\%$ confidence” means that $95\%$ confidence of interval created from samples will contain the true proportion.