Students enrolled in an introductory statistics course at a

Students enrolled in an introductory statistics course at a university were asked to take a survey that indicated whether the student has a visual or verbal learning style. Of the 39 students who took the survey, 25 were judged to have a visual learning style and 14 were considered verbal learners.

1. Determine a 90% confidence interval for the population proportion who are visual learners at this university.

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2. Write a sentence interpreting what this interval says.

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3. How would a 99% confidence interval compare to this one in terms of its midpoint and half-width? (Do not bother to determine this interval.)

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4. Check whether the technical condition concerning sample size is satisfied here.

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5. Explain why you might feel wary about applying this confidence interval to the population of all students at this university.

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Solution

1) a 90% C.I. = 25/39 +- 1.28 *Sqrt(25/39 *(1-25/39)/39) = (0.542, 0.739)
2) With 90% confidence the interval (0.542, 0.739) contains the population proportion of visual learners.
3) a 99% C.I. would have the same midpoint but would be wider.
4) (Depends on your book) n > 30 and np > 5 and n(1-p) > 5 are all satisfied here.
5) Theres some serious sampling bias here if we try to apply these results to all university students. First, it was a statistics class and there is no guarantee that students in statistics class have the same learning style as those in other classes. Second, students who agree to do this kind of thing are more compliant and obedient than other people which may affect learning style. Third, there could have been 170 people signed up for the class. If only 39 people showed up for class to take this stupid survey, the other 131 people probably had a learning style that didn\'t involve going to class.