Chapter 20, Problem 17E

(a)

To determine

To find out what are the hypotheses.

Answer to Problem 17E

  $\begin{array}{l}{H}_0:p=0.63\\ H_a:p>0.63\end{array}$

Explanation of Solution

It is given in the question that according to law school, $63\%$ of law school applicants were accepted to some law school. The training program claims that $163\text{ of the 240}$ students trained were admitted to law school. Thus, the null hypothesis will be that the rate will be equal to $63\%$ . And the alternative hypothesis is that the rate will be greaterthan $63\%$ . So, we define hypotheses as:

  $\begin{array}{l}{H}_0:p=0.63\\ H_a:p>0.63\end{array}$

(b)

To determine

To check the conditions and find the P-value.

Answer to Problem 17E

Yes, all the assumptions and conditions for inference met and the $P-\text{value}=0.057$ .

Explanation of Solution

It is given in the question that according to law school, $63\%$ of law school applicants were accepted to some law school. The training program claims that $163\text{ of the 240}$ students trained were admitted to law school. So, we define hypotheses as:

  $\begin{array}{l}{H}_0:p=0.63\\ H_a:p>0.63\end{array}$

Now, 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 population.

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

  $\begin{array}{l}np=0.63(240)>10\\ n(1-p)=0.37(240)>10\end{array}$

Thus, all the conditions are satisfied.Thus, for this first we have to select $5:1-\text{PropZInt}$ from the STAT TESTS menu. Then, enter the observed count and sample size for the sample. 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,

  $\begin{array}{l}x=151.2\\ n=240\\ c=0.95=95\%\end{array}$

  $\begin{array}{l}{H}_0:p=0.63\\ H_a:p>0.63\end{array}$

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

The value of z-statistics and the P-value will be then,

  $\begin{array}{l}z=1.578\\ P-\text{value}=0.057\end{array}$

(c)

To determine

To explain would you recommend this program based on what you see here.

Answer to Problem 17E

No, we will not recommend.

Explanation of Solution

It is given in the question that an organization send free mailing labels and greeting cards to potential donors on their list and ask for a voluntary contribution. They have had a contribution rate of $5\%$ in past campaigns, but the staff worries that the rate will be lower. So, we define hypotheses as:

  $\begin{array}{l}{H}_0:p=0.05\\ H_a:p<0.05\end{array}$

The value of z-statistics and the P-value will be then,

  $\begin{array}{l}z=1.578\\ P-\text{value}=0.057\end{array}$

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 there is no sufficient evidence to show that the program demonstrate real improvement. No, we will not recommend this program.