Chapter PV, Problem 31RE

(a)

To determine

To create a $95\%$ confidence interval for the difference in graduation rates between children of normal and children of very low birth weights and check the assumptions and conditions.

Answer to Problem 31RE

  $(0.01516,0.16215)$ .

Explanation of Solution

Let us first check the conditions and assumptions for this distribution as:

Random condition: It is satisfied as children are randomly selected from the population.

Independent condition: It is satisfied as groups do not affect each other.

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

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

  $\begin{array}{l}np=193>10\\ n(1-p)=40>10\end{array}$

  $\begin{array}{l}np=179>10\\ n(1-p)=63>10\end{array}$

Thus, all the conditions are satisfied. Therefore, now, to find the confidence interval we will use the calculator $\text{TI}-89$ . So, we have to first go to the STAT $\text{Ints}$ menu and select $6:2-\text{PropZInt}$ . Then we have to enter the observed counts and the sample sizes for both the samples. Then specify a confidence level. Then the interval will be calculated. Thus, we have,

  $\begin{array}{l}{x}_1=193\\ n_1=233\\ x_2=179\\ n_2=242\\ \alpha =0.05\end{array}$

Thus, the confidence interval will be as:

  $(0.01516,0.16215)$ .

(b)

To determine

To explain does this provide evidence that premature birth may be a risk factor for not finishing high school and use your confidence interval to test an appropriate hypothesis.

Answer to Problem 31RE

There is evidence that premature birth decreases the probability of finishing high school.

Explanation of Solution

It is given in the question that,

  $\begin{array}{l}{x}_1=193\\ n_1=233\\ x_2=179\\ n_2=242\\ \alpha =0.05\end{array}$

Thus, the confidence interval will be as:

  $(0.01516,0.16215)$ .

Thus, to calculate an appropriate hypothesis we will use $\text{TI}-89$ calculator. Thus, for this first we have to select $6:2-\text{PropZInt}$ from the STAT TESTS menu. Then, enter the observed counts and sample sizes for both the samples. Then we will indicate what kind of test we want: one-tail lower tail, two-tail, or one-tail upper tail. Then, we will specify whether to calculate the result or draw the result. Hence we have,

  $\begin{array}{l}{x}_1=193\\ n_1=233\\ x_2=179\\ n_2=242\\ \alpha =0.05\\ H_0:p_1=p_2\\ H_a:p_1<p_2\end{array}$

So, by putting the values above by using the calculator we will get the result as,

The P-value will be then,

  $P=0.00954$

And as we know if the P-value is smaller than the significance level, then reject the null hypothesis, so, we have,

  $P<0.05\Rightarrow \text{ Reject }{H}_0$

Thus, we conclude that there is evidence that premature birth decreases the probability of finishing high school.

(c)

To determine

To find which type of error did you make, if your conclusion is incorrect.

Answer to Problem 31RE

It is Type-I error.

Explanation of Solution

It is given in the question that,

  $\begin{array}{l}{x}_1=193\\ n_1=233\\ x_2=179\\ n_2=242\\ \alpha =0.05\end{array}$

Thus, the confidence interval will be as: $(0.01516,0.16215)$ . And as we know if the P-value is smaller than the significance level, then reject the null hypothesis, so, we have,

  $P<0.05\Rightarrow \text{ Reject }{H}_0$

Thus, we conclude that there is evidence that premature birth decreases the probability of finishing high school. Thus, if we are rejecting the null hypothesis if it is true as we are supposing that the conclusion is wrong, then it is Type-I error.