Question

A friend of mine who works at Colorado State University (CSU) claimed that CSU students are more likely than CU students to skip class on powder days (i.e., days following an amazing snowfall in the mountains). Last year, we tracked attendance in our classes across two weeks and counted the number of absent students that occurred on powder days. We compare our data in the table below: Skipped Class? Yes No CU-Boulder 34 140 CSU 28 80 a. What proportion of CU students skipped class on powder days? What proportion of CSU students skipped class on powder days? b. State the research and null hypotheses to test my friend's claim. c. Using an alpha level of .01, determine the critical value: d. What is the obtained value for the test? e. From this test, what should I conclude about my friend's claim?

Answer 1

a. Proportion of CU students who skipped class on powder days = 34/(34+140) ≈ 0.195 or 19.5%

Proportion of CSU students who skipped class on powder days = 28/(28+80) ≈ 0.259 or 25.9%

b. Research hypothesis: The proportion of CSU students who skip class on powder days is greater than the proportion of CU students who skip class on powder days.

Null hypothesis: The proportion of CSU students who skip class on powder days is not greater than or equal to the proportion of CU students who skip class on powder days.

c. Degrees of freedom = (2-1)*(2-1) = 1

Using an alpha level of .01 and the chi-squared distribution table with 1 degree of freedom, the critical value is 6.63.

d. Observed value of the test statistic = [(34/(34+140))-(28/(28+80))]^2 / [(34+140+28+80)/((34+140)*(28+80))] ≈ 1.41

e. Since the observed value of the test statistic (1.41) is less than the critical value (6.63), we fail to reject the null hypothesis. Therefore, we do not have sufficient evidence to support my friend's claim that CSU students are more likely to skip class on powder days than CU students.

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