One criticism of reading comprehension tests is that while they may measure reading comprehension, they also measure other factors not related to reading comprehension. A reading comprehension (RC) test on a major college entrance exam provides short English prose passages, and the examinees answer a set of multiple-choice items about the passage. To see if particular items measure something other than RC, investigators gave the RC test without the reading passages to a random sample of psychology students. 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).
Suppose that on one question, 30 out of 100 examinees answered the question correctly. Is this sufficient evidence that students are using more than just reading comprehension to answer this question? Test the relevant hypothesis using alpha = .05

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

[tex]H_0: p = 0.2[/tex]

At the alternate hypothesis, we test that the proportion is above 20%, that is:

[tex]H_a: p > 0.2[/tex]

The test statistic is:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.

0.2 is tested at the null hypothesis:

This means that [tex]\mu = 0.2, \sigma = \sqrt{0.2*0.8} = 0.4[/tex]

Suppose that on one question, 30 out of 100 examinees answered the question correctly.

This means that [tex]n = 100, X = \frac{30}{100} = 0.3[/tex]

Test statistic:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{0.3 - 0.2}{\frac{0.4}{\sqrt{100}}}[/tex]

[tex]z = 2.5[/tex]

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