Question

A true/false test has 40 questions. Suppose a passing grade is 30 or more correct answers. Test the claim that a student knows more than half of the answers and is not just guessing. Assume the student gets 30 answers correct out of 40. Use a significance level of 0.05. Steps 1 and 2 of a hypothesis test procedure are given below. Show step 3: finding the test statistic and the p-vahie and step 4, interpreting the results. Step 1:H0: p = 0.50 Ha: p > 0.50 Step 2: Choose the one-proportion z-test. Sample size is large enough, because np0 is 40(0.5) = 20 and n(1 - p0) = 40(.5) = 20, and both are more than 10. Assume the sample is random. Step 3: Compute the z-test statistic, and the p-value. z = (Round to two decimal places as needed.) p-value = Q (Round to three decimal places as needed.)

Answer 1

Answer:

z≈3.16

p≈0.001

we reject the null hypothesis and conclude that the student knows more than half of the answers and is not just guessing in 0.05 significance level.

Step-by-step explanation:

As a result of step 2, we can assume normal distribution for the null hypothesis

step 3:

z statistic is computed as follows:

z=$\frac{X-M}{\frac{\sqrt{p*(1-p)} }{\sqrt{N} } }$ where

• X is the proportion of correct answers in the test ($\frac{30}{40}$)
• M is the expected proportion of correct answers according to the null hypothesis (0.5)

• p is the probability of correct answer (0.5)
• N is the total number of questions in the test (40)

z=$\frac{0.75-0.50}{\frac{\sqrt{0.5*(0.5)} }{\sqrt{40} } }$ ≈ 3.16

And corresponding p value for the z-statistic is p≈0.001.

Since p<0.05, we reject the null hypothesis and conclude that the student knows more than half of the answers and is not just guessing in 0.05 significance level.