students enrolled non an introductory statics course at a university were ask to

Question

students enrolled non an introductory statics course at a university were ask to take a survey that indicate whether the students has a visual learning or verbal learning style.of the 39 students who took the survey, 25 were judge to have learning style and 14 were considered verbal learners.

determine a 90% confidence intervals for the population proportion who are visual learners at this university?

write a statement interrupting what the interval is saying?

How would a 99% confidence interval compare to this one in terms of its midpoints and half width?

check whether the technical conditions concerning sample size is satisfied here

Explanation / Answer

Solution:

- A 90% Confidence Interval

p ± Z/2 * sqrt[p(1-p)/n]

= 25/39 +- 1.28 *Sqrt[25/39 *(1-25/39)/39]
= (0.542, 0.739)
- With 90% confidence the interval (0.542, 0.739) contains the population proportion of visual learners.
-A 99% Confidence Interval would have the same midpoint but would be wider.
- n > 30 and np > 5 and n(1-p) > 5 are all satisfied here.

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