Question

Literacy A researcher reports that $80$% of high school graduates but only $40$% of high school dropouts would pass a basic literacy test.^$5$ Assume that the researcher’s claim is true. Suppose we give a basic literacy test to a random sample of $60$high school graduates and a separate random sample of $75$high school dropouts.

(a) Find the probability that the proportion of graduates who pass the test is at least $0.20$higher than the proportion of dropouts who pass. Show your work.

(b) Suppose that the difference in the sample proportions (graduate – dropout) who pass the test is exactly $0.20$. Based on your result in part (a), would this give you a reason to doubt the researcher’s claim? Explain.

Short answer

From the given information,

a) 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) Yes, the claim of the researcher should be questioned.

Step-by-step solution

Part(a) Step 1: Given Information

It is given in the question that,

The proportion of high school graduates that will pass the basic literacy test=$80$%

The 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 =$75$

Part(a) Step 2: Explanation

Let $p1$be the percentage of high school graduates who pass a basic literacy exam, and $p2$be the proportion of high school dropouts who pass a basic literacy test. $p1-p2$'s sampling distribution is Normal because:

localid="1650446649236" $$\begin{gathered} n_1{p}_1=60\times 0.80 \\ =48 \end{gathered}$$

localid="1650446659437" $$\begin{gathered} n_1(1-p_1)=60\times 0.20 \\ =12 \end{gathered}$$

localid="1650446668572" $$\begin{gathered} n_2{p}_2=75\times 0.40 \\ =30 \end{gathered}$$

localid="1650446679632" $$\begin{gathered} n_2(1-p_2)=75\times 0.60 \\ =45 \end{gathered}$$

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

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

$=0.80-0.40$

$=0.40$

Part (a) Step 3: Explanation

The standard deviation is:

${\sigma }_{i-{\dot{p}}_2}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}$

$=\sqrt{\frac{0.8\times (1-0.8)}{60}+\frac{0.40\times (1-0.40)}{75}}$

$=0.0766$

Hence,

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

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

$=1-P(z\le -2.61)$

$=1-0.00453$

$=0.9955$

Part(b) Step 1: Given Information

It is given in the question that,

The proportion of high school graduates that will pass the basic literacy test=$80$%

The 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 =$75$

Part (b) Step 5: Explanation

Yes, the claim of the researcher should be questioned. The probability that at least $0.20$or $20$% more graduates pass the test than dropouts pass is $0.9955$. There is a $99.55$% likelihood of receiving samples with at least $20$% more high school graduates passing, but only a $0.45$percent risk of receiving samples with no more than $20$% more high school graduates passing.