Chapter 10.1, Problem 6E

(a)

To determine

The probability that the proportion of graduates passing the test is at least 0.20 higher than the proportion of dropouts passing.

Answer to Problem 6E

The probability that the proportion of red jelly beans in the Child sample is less than or equal to the proportion of red jelly beans in the Adult sample is $\mathrm{0.0212.}$

Explanation of Solution

Given information:

Proportion of high school graduates that will pass basic literacy test = 80%

Proportion of high school dropouts that will pass basic literacy test = 40%

Number of Random Sample from high school graduates for basic literacy test = 60

Number of Random Sample from high school dropouts for basic literacy test = 75

Let $p_1$ is the actual proportion of high school graduates who pass a basic literacy test and $p_2$ is the actual proportion of high school dropouts who pass a basic literacy test. The sampling distribution of $p_1-p_2$ is Normal because:

  $\begin{array}{cc}{n}_1{p}_1& =60\times 0.80=48\\ n_1\left(1-p_1\right)& =60\times 0.20=12\\ n_2{p}_2& =75\times 0.40=30\\ n_2\left(1-p_2\right)& =75\times 0.60=45\end{array}$

Are at least $10,$ the sampling distribution of ${\widehat{p}}_1-{\widehat{p}}_2$ is approximately Normal. Its mean is:

  ${\mu }_{\dot{A}-{\dot{p}}_2}=p_1-p_2$

  $\begin{array}{l}=0.80-0.40\hfill \\ =0.40\hfill \end{array}$

The standard deviation is:

  $\begin{array}{cc}{\sigma }_{\text{<mover accent="true"> <mo>A </mo><mo>˙</mo></mover>}-{\dot{p}}_2}& =\sqrt{\frac{p_1\left(1-p_1\right)}{n_1}+\frac{p_2\left(1-p_2\right)}{n_2}}\\ & =\sqrt{\frac{0.8\times (1-0.8)}{60}+\frac{0.40\times (1-0.40)}{75}}\\ & =0.0766\end{array}$

Hence,

  $p\left({\widehat{p}}_1-{\widehat{p}}_2\ge 0.2\right)=p\left(z\ge \frac{0.2-0.40}{0.0766}\right)$

  $=P(z\ge -2.61)$ (From normal table)

  $\begin{array}{l}=1-P(z\le -2.61)\hfill \\ =1-0.00453\hfill \\ =0.9955\hfill \end{array}$

The probability that the proportion of graduates who pass the test is at least 0.20 higher than the proportion of dropouts who pass is 0.9955

(b)

To determine

To Explain: The difference in the sample proportions (graduate - dropout) who pass the test is exactly $\mathrm{0.20.}$ Based on result in part (a), can the researcher's claim should be doubted.

Answer to Problem 6E

Yes the company claim can be doubted.

Explanation of Solution

Yes, the researcher's claim should be doubted. The probability that the proportion of graduateswho pass the test is at least 0.20 or $20\text{\%}$ higher than the proportion of dropouts who pass is $\mathrm{0.9955.}$ There is a $99.55\text{\%}$ chance of getting samples where at least $20\text{\%}$ more of the highschool graduates pass, there is only a $0.45\text{\%}$ chance of getting a sample where no more than $20\text{\%}$ more of the high school graduates pass.