Question 2: Last semester, Oklahoma State social media had many messages express

Question

Question 2: Last semester, Oklahoma State social media had many messages expressing discontent with the guest speaker’s $90,000 cost. Say student government decides to conduct a poll about activity fee money. They survey an independent random sample of 55 Oklahoma State students. Of the surveyed students, 23 said that they would prefer spending $90,000 on bouncy houses instead of guest speakers.

[5 points] Compute a 98% confidence interval for the population proportion of students that would prefer bouncy houses over guest speakers.

[3] Based on the interval you computed, is it credible to claim that a majority (more than 50%) of students would prefer bouncy houses? Why or why not?

Explanation / Answer

Solution:

Given n = 55
p = 23/25 = 0.42
The 98% confidence interval for population proportion of student is
  
= p ± Z/2.sqrt[p(1-p)/n]
= 0.42 ± Z0.02/2 . sqrt[0.42(1-0.42)/25]
= 0.42 ± 2.58 * sqrt[0.42(1-0.42)/25]
= [ 0.1653, 0.6747]
The proportion of the student are lies in between 0.1653 and 0.6747 at = 0.02

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