Answer:
The p-value of the test is 0.0062 < 0.05, which means that this is sufficient evidence that students are using more than just reading comprehension to answer this question
Step-by-step explanation:
The investigators reasoned that if questions were measuring knowledge or memory rather than just RC, students would answer questions at a higher rate than chance (20%, since there were 5 choices for each question).
This means that at the null hypothesis, we test that the proportion is the probability of answering correctly by change, that is 20%. So
![H_0: p = 0.2](https://tex.z-dn.net/?f=H_0%3A%20p%20%3D%200.2)
At the alternate hypothesis, we test that the proportion is above 20%, that is:
![H_a: p > 0.2](https://tex.z-dn.net/?f=H_a%3A%20p%20%3E%200.2)
The test statistic is:
![z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
In which X is the sample mean,
is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
0.2 is tested at the null hypothesis:
This means that ![\mu = 0.2, \sigma = \sqrt{0.2*0.8} = 0.4](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.2%2C%20%5Csigma%20%3D%20%5Csqrt%7B0.2%2A0.8%7D%20%3D%200.4)
Suppose that on one question, 30 out of 100 examinees answered the question correctly.
This means that ![n = 100, X = \frac{30}{100} = 0.3](https://tex.z-dn.net/?f=n%20%3D%20100%2C%20X%20%3D%20%5Cfrac%7B30%7D%7B100%7D%20%3D%200.3)
Test statistic:
![z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
![z = \frac{0.3 - 0.2}{\frac{0.4}{\sqrt{100}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B0.3%20-%200.2%7D%7B%5Cfrac%7B0.4%7D%7B%5Csqrt%7B100%7D%7D%7D)
![z = 2.5](https://tex.z-dn.net/?f=z%20%3D%202.5)
P-value of the test and decision:
The p-value of the test is the probability of finding a sample proportion above 0.3, which is 1 subtracted by the p-value of z = 2.5.
Looking at the z-table, z = 2.5 has a p-value of 0.9938
1 - 0.9938 = 0.0062
The p-value of the test is 0.0062 < 0.05, which means that this is sufficient evidence that students are using more than just reading comprehension to answer this question