Chapter PVII, Problem 19RE

(a)

To determine

To test the appropriate hypotheses, using the two-proportion z -procedure.

Answer to Problem 19RE

There is evidence of a difference in effectiveness.

Explanation of Solution

It is given that,

  $\begin{array}{l}{x}_1=42\\ n_1=157\\ x_2=7\\ n_2=89\end{array}$

Let us assume that all the conditions for inference are met. Thus, to do the mechanics of a hypothesis test for equality of population proportions we will use the calculator $\text{TI}-89$ . Then select $6:2-$ PropZTestfrom the STAT Tests menu. Enter the observed counts and sample sizes. Indicate what kind of test you want: one-tail upper tail, lower tail, or two-tail. And specify whether results should simply be calculated or displayed with the area corresponding to the P-value of the test shaded. The hypothesis is defined as:

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_0:p_1\ne p_2\end{array}$

Thus, by using the calculator $\text{TI}-89$ , the test statistics and the P-value is as:

  $\begin{array}{l}z=3.56\\ P=0.0004\end{array}$

As we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, so we have,

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

Thus, we conclude that there is evidence of a difference in effectiveness and it appears the methods are not as good as older women.

(b)

To determine

To repeat the analysis, using an appropriate chi-square procedure.

Answer to Problem 19RE

Explanation of Solution

It is given that,

  $\begin{array}{l}{x}_1=42\\ n_1=157\\ x_2=7\\ n_2=89\end{array}$

Let us assume that all the conditions for inference are met.The contingency table is as follows:

Age	Successes	failure	Total
under age 38	42	115	157
38 and older	7	82	89
Total	49	197	246

Let us assume that all the conditions for inference are met.Thus, to test a hypothesis of homogeneity or independence, we will use the calculator $\text{TI}-89$ so, we need to enter the data as a matrix. From the home screen, press APPS and select $6:$ Data/Matrix Editor, then select $3:$ New. Specify type as Matrix and name the matrix in the Variable box. Specify the number of rows and columns. Type the entries, pressing ENTER after each. Press $2\text{nd ESC}$ to leave the editor. To do the test, choose $\text{8:Chi2 2-way}$ from the STAT Tests menu.The hypothesis is defined as:

Null hypothesis: There is no difference in effectiveness.

Alternative hypothesis: There is difference in effectiveness.

Thus, by using the calculator $\text{TI}-89$ , the test statistics and the P-value is as:

  $\begin{array}{l}{\chi }^2=12.70\\ P=0.0004\end{array}$

As we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, so we have,

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

Thus, we conclude that there is evidence of a difference in effectiveness and it appears the methods are not as good as older women.

(c)

To determine

To explain how the two results are equivalent.

Explanation of Solution

It is given that,

  $\begin{array}{l}{x}_1=42\\ n_1=157\\ x_2=7\\ n_2=89\end{array}$

Thus, from both the tests in part (a) and part (b), we conclude that there is evidence of a difference in effectiveness and it appears the methods are not as good as older women. So, this two results are equivalent as there P-value is equal and also,

  ${\chi }^2=12.70={(3.563944)}^2=z^2$