According to U.S. News and World Reports, 18.60% of a certain online school stud

Question

According to U.S. News and World Reports, 18.60% of a certain online school students have military experience. In a random sample of 40 students enrolled in the online program of STAT 101, 5 reported having military experience.

1.) If the population proportion is 18.60%, what is the probability of taking a random sample of n=40 and finding a sample proportion more extreme than the one observed in this sample? Use StatKey (http://www.lock5stat.com/StatKey/) to determine this proportion. Include a screenshot of your sampling distribution with this proportion highlighted.

2.) Given your results from part D, do you think that the proportion of all STAT 200 students who have military experience is different from the overall population of World Campus students where p=0.1860? Explain your reasoning.

Explanation / Answer

1) p^ = 5/40 = 1/8 = 0.125

here p = 0.186

Z = (p^ - p)/sqrt(pq/n)

P(p^ < 0.125)

P(Z < (0.125 -0.186) /sqrt(0.186*814/40))

P(Z < -0.0313538)

=0.4875000

2) p-value

= 2 P(p^ < 0.125) = 2*0.4875 = 0.975

since p-value > 0.05 (level of significance)

we fail to reject the null hypothesis and conclude that there is no significant evidence that the proportion of all STAT 200 students who have military experience is different from the overall population of World Campus students where p=0.1860

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