Chapter 20, Problem 18E

(a)

To determine

To find out what are the hypotheses.

Answer to Problem 18E

  $\begin{array}{l}{H}_0:p=0.46\\ H_a:p<0.46\end{array}$

Explanation of Solution

It is given in the question that according to medical colleges, only $46\%$ of medical school applicants were admitted to a medical school. The college expressed concerns that only $77\text{ of the 180}$ students in their class who applied to medical school were admitted. Thus, the null hypothesis will be that the rate will be equal to $46\%$ . And the alternative hypothesis is that the rate will be lessthan $46\%$ . So, we define hypotheses as:

  $\begin{array}{l}{H}_0:p=0.46\\ H_a:p<0.46\end{array}$

(b)

To determine

To check the conditions and find the P-value.

Answer to Problem 18E

Yes, all the assumptions and conditions for inference met and the $P-\text{value}=0.193$ .

Explanation of Solution

It is given in the question that according to medical colleges, only $46\%$ of medical school applicants were admitted to a medical school. The college expressed concerns that only $77\text{ of the 180}$ students in their class who applied to medical school were admitted. So, we define hypotheses as:

  $\begin{array}{l}{H}_0:p=0.46\\ H_a:p<0.46\end{array}$

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,

  $\begin{array}{l}np=0.46(180)>10\\ n(1-p)=0.54(180)>10\end{array}$

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,

  $\begin{array}{l}x=82.8\\ n=180\\ c=0.95=95\%\\ H_0:p=0.46\\ H_a:p<0.46\end{array}$

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,

  $\begin{array}{l}z=-0.867\\ P-\text{value}=0.193\end{array}$

(c)

To determine

To explain are the trustees right to be concerned or is the president correct.

Answer to Problem 18E

The high p-value indicates that this is not an unusual result and may just be due to year-to-year variation as the president says.

Explanation of Solution

It is given in the question that according to medical colleges, only $46\%$ of medical school applicants were admitted to a medical school. The college expressed concerns that only $77\text{ of the 180}$ students in their class who applied to medical school were admitted. So, we define hypotheses as:

  $\begin{array}{l}{H}_0:p=0.46\\ H_a:p<0.46\end{array}$

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

  $\begin{array}{l}z=-0.867\\ P-\text{value}=0.193\end{array}$

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 the high p-value indicates that this is not an unusual result and may just be due to year-to-year variation as the president says.