Question

The university is concerned about a rise in the number of academic offences committed by undergraduate students. They believe that students do not understand the seriousness of such offences. They develop a course that outlines the potential consequences for students who are involved in academic offenses (e.g. cheating, plagiarism, etc.) in order to deter students’ acceptability of committing these offenses. All first year students are required to take the course. To assess the effectiveness of the course in deterring (or reducing) acceptability of academic offences, before the course starts, a random sample of 150 first year students are asked to complete a questionnaire that includes a scale to assess their views on the acceptability of committing an academic offence. After the course is over, a different sample of 156 first year students complete the same questionnaire. Scores on the scale range from 0 to 25 (interval/ratio measure), with a mean (X") and standard deviation (s) determined for the scores in each sample. Higher scores reflect GREATER acceptability of academic offences.

(a) What are the null hypothesis and alternate or research hypothesis that you would state in order to assess the effectiveness of the course in deterring acceptability of academic offences?

(b) What sampling distribution would you use, and what formula(s) would you use?

(c) Given an alpha level of 0.05, what is the critical value of the test statistic?

(d) If the value of the obtained test statistic is +1.754, what would you conclude about the effectiveness of the course? Be sure to state the conclusion in statistical terms, and in words.

Show your work

Answer 1

a)

Answer:

Null hypothesis: The mean score in the questionnaire after the course end is equal to the mean score in the questionnaire before the course start.

Alternative hypothesis: The mean score after the course end is higher compared to the mean score before the course start.

Explanation:

The null hypothesis is defined as the mean score in the questionnaire before the course start and after the course end are equal. And the alternative hypothesis tests the claim that the mean score in the questionnaire after the course end is significantly higher than the mean score in the questionnaire before the course start.

b)

Answer:

The sample distribution of the sample means will be used.

Sampling distribution of the difference in means,

Explanation:

Since we are comparing the sample mean score in the questionnaire, the sample distribution of the sample means will be used.

Since the population standard deviation is not known, the two-sample t-test is used to test the hypothesis.

Test statistic,

The t statistic is obtained using the formula,

c)

Answer:

t critical value = 1.6499

Explanation:

Since the population standard deviation is not known, the t distribution is used to test the hypothesis.

The t critical value is obtained from the t critical value table for significance level = 0.05 and the degree of freedom = n1+n2-2=150+156-2=304 and for the right-tailed test.

d)

Answer:

The course effectively increases the acceptability of committing an academic offense

Explanation:

test statistic, t = 1.754

critical value, tcritical = 1.6499

Since the t statistic is greater than the t critical value, the null hypothesis is rejected hence there is sufficient evidence to conclude that the mean score in the questionnaire after the course end is higher compared to the mean score in the questionnaire before the course start.

This means the course effectively increases the acceptability of committing an academic offense