d. An article online claims that 17% of all World Campus students are over the a

Question

d. An article online claims that 17% of all World Campus students are over the age of 40. What is the probability that a population where p=.17 would produce a random sample of n=120 with a sample proportion as high (or higher) than the one found in this sample?

In other words, given that the population proportion (p) is .17, what proportion of samples of n=120 would have a sample proportion greater than the one that you observed in your sample?

You will need to compute the standard error again because now we have a population parameter: p=.17

Given your results from part (d), do you think that the article’s claim that the population proportion is .17 could be accurate? Or, do you think that their claim is an underestimate of the true proportion of World Campus students who are over the age of 40? Explain your reasoning.

Explanation / Answer

d.

Proportion ( P ) =0.17
Standard Deviation ( sd )= Sqrt (P*Q /n) = Sqrt(0.17*0.83/120)
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
              
P(X > 0.2) = (0.2-0.17)/0.0343
= 0.03/0.0343 = 0.8746
= P ( Z >0.875) From Standard Normal Table
= 0.1909                  
                  
approx 19.09% are produce higher than 0.20

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