Chapter PV, Problem 17RE

(a)

To determine

To write an appropriate hypothesis.

Answer to Problem 17RE

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1<p_2\end{array}$

Explanation of Solution

It is given in the question about the disease eclampsia. Thus, the null hypothesis is that there is no difference between the treatment and eclampsia and the alternative hypothesis is that there is difference i.e. treatment prevents deaths from eclampsia. So, we define hypotheses as:

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1<p_2\end{array}$

(b)

To determine

To check the assumptions and conditions.

Answer to Problem 17RE

All are satisfied.

Explanation of Solution

It is given in the question about the disease eclampsia. So, we define hypotheses as:

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1<p_2\end{array}$

Thus, let us check the conditions and assumptions necessary for inference as:

Random condition: It is satisfied as 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 the sample is less than $10\%$ of all people with Eclampsia.

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

For sulphate:

  $\begin{array}{l}np=11>10\\ n(1-p)=29>10\end{array}$

For Placebo:

  $\begin{array}{l}np=20>10\\ n(1-p)=76>10\end{array}$

Thus, all the conditions are satisfied.

(c)

To determine

To find the P-value of the test.

Answer to Problem 17RE

  $P=0.8008$ .

Explanation of Solution

It is given in the question about the disease eclampsia. So, we define hypotheses as:

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1<p_2\end{array}$

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,

  $\begin{array}{l}{x}_1=11\\ n_1=40\\ x_2=20\\ n_2=96\\ \alpha =0.05\\ H_0:p_1=p_2\\ H_a:p_1<p_2\end{array}$

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

The P-value will be then,

  $P=0.8008$

(d)

To determine

To explain what do you conclude about the magnesium sulphide treatment.

Explanation of Solution

It is given in the question about the disease eclampsia. So, we define hypotheses as:

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1<p_2\end{array}$

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,

  $\begin{array}{l}{x}_1=11\\ n_1=40\\ x_2=20\\ n_2=96\\ \alpha =0.05\\ H_0:p_1=p_2\\ H_a:p_1<p_2\end{array}$

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

The P-value will be then,

  $P=0.8008$

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

  $P>0.05\Rightarrow \text{ Fail to Reject }{H}_0$

Thus, we conclude that here is no evidence that it helps prevent the women’s death.

(e)

To determine

To explain which type of error have you made if your conclusion is wrong.

Answer to Problem 17RE

It is Type II error.

Explanation of Solution

It is given in the question about the disease eclampsia. So, we define hypotheses as:

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1<p_2\end{array}$

The P-value will be then,

  $P=0.8008$

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

  $P>0.05\Rightarrow \text{ Fail to Reject }{H}_0$

Thus, we conclude that here is no evidence that it helps prevent the women’s death. Thus, if our conclusion is wrong then it is Type II error we have made.

(f)

To determine

To name two things you could do to increase the power of this test.

Explanation of Solution

It is given in the question about the disease eclampsia. So, we define hypotheses as:

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1<p_2\end{array}$

The P-value will be then,

  $P=0.8008$

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

  $P>0.05\Rightarrow \text{ Fail to Reject }{H}_0$

Thus, we conclude that here is no evidence that it helps prevent the women’s death. Thus, to increase the power of test we will increase the sample size as it decreases sample variation but is costly. And we will increase $\alpha$ as it decreases chance of Type II error but increases chance of Type I error.

(g)

To determine

To find out what are the advantages and disadvantages of those two options.

Explanation of Solution

It is given in the question about the disease eclampsia. So, we define hypotheses as:

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1<p_2\end{array}$

The P-value will be then,

  $P=0.8008$

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

  $P>0.05\Rightarrow \text{ Fail to Reject }{H}_0$

Thus, we conclude that here is no evidence that it helps prevent the women’s death. Thus, to increase the power of test we will increase the sample size as it decreases sample variation but is costly. And we will increase $\alpha$ as it decreases chance of Type II error but increases chance of Type I error.