A professor creates two versions of a 50-question multiple-choice quiz. Each que

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A professor creates two versions of a 50-question multiple-choice quiz. Each question has four choices. One student got a soore of 45 out of 50 for the version of the test given to the person siting next to him. The professor thinks the student was copying another exam. The student admits that he hadn't studied for the test, but he says he was simply guessing on each question and just got lucky. For the professor chooses the comect answer if just guessing, and the alternative is p>0.25 Would the p-value for this hypothesis test be high or low? Explain , the null hypothesis is that p 025, where p is the probability that the student if the student were would be | | An olcome has occurred that would be zes ten Click to select your answerfs). Sample Tests and Quizzes

Explanation / Answer

Solution:- The pbe -value would be almost 0. An outcome has occured that would be very uncertain if student were simply guessing.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P < 0.25

Alternative hypothesis: P > 0.25

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).

= sqrt[ P * ( 1 - P ) / n ]

= 0.06124

z = (p - P) /

z = 10.6

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is more than 10.6. We use the Normal Distribution Calculator to find P(z > 10.6).

Thus, the P-value = less than 0.0001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we cannot accept the null hypothesis.

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