🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning

Like

Report

Numerade Educator

Like

Report

Problem 32 Hard Difficulty

The Cincinnati Enquirer reported that, in the United States, 66$\%$ of adults and 87$\%$ of youths ages 12 to 17 use the Internet (The Cincinnati Enquirer, February $7,2006 ) .$ Use the reported numbers as the population proportions and assume that samples of 300 adults and 300 youths will be used to learn attitudes toward Internet security.
a. Show the sampling distribution of $\overline{p},$ where $\overline{p}$ is the sample proportion of adults using the Internet.
b. What is the probability that the sample proportion of adults using the Internet will be within $\pm .04$ of the pooulation proportion?
c. Whatis the probabilit that the sample proportion of youths using the Internet will be within $\pm .04$ of the population proportion?
d. Is the probability different in parts (b) and (c)? If so, why?
e. Answer part (b) for a sample of size $600 .$ Is the probability smaller? Why?

Answer

a. See graph
b. 0.8558
c. 0.9606
d. Different, because the population proportions are different.
e. 0.9616

Discussion

You must be signed in to discuss.

Video Transcript

all right, So this question gives us population proportions of adult and youth populations, and it asks us to compute some sampling distribution probabilities. So the first part wants us to draw the distribution of P hats for the adults. So since we're dealing with a large sample, we can guarantee that it's normal based on our large counts condition. So we can just draw a bell curve here with a mane at the population proportion and a sigma of square root 0.66 0.34 over 300 based on our formula. So now, for Part B, it wants us to compute the probability that a sample of 300 adults is within point zero for so it wants probability that we're within 0.4 which is the same thing is asking probability that are lower bound for P hat, his 0.62 and our upper bound it is 0.70 And this, of course, is a normal CDF question. So we can write that normal CDF are lower bound is 0.62 Our upper bound is 0.7. Our mean is 0.66 and our standard air is the square root of our population proportion times one, minus the population proportion all over the square root of our sample size. And if you put that into your calculator that works out to be 0.8564 then it wants the same question except for the youth. So probability that we're within 0.4 which in this case is the same, is asking probability that we're between 0.83 and 0.91 which is another normal CDF are lower bound. This 0.83 Our upper bound is 0.91 Our mean is 0.87 and our standard air. It's different 2.87 times 0.13 all divided by the square root of our sample size, which in this case works out to be point 9606 Now it asks us to compare these two answers. So we have 0.850 point 8564 versus 85640.9606 and these are not the same. There's actually quite a bit of difference here, so it's asking why is there a difference? So we can say that the probabilities are different are different because the standard error is different. So if we look, standard error is given by the square root of P Q over end, and in this case, our sample size is the same. But our P and Q are different, so they're really different because we have different P. The population proportions air different for each, which causes are probabilities not to line up. Then par EE wants us to repeat the calculation for adults, but this time bumping up the sample size to 600. So we're asking the exact same question, the probability that were between 0.62 and 0.70 So again, normal CDF are lower bound 0.62 our upper bound this 0.70 our mean is 0.66 and our standard error is what changes this time 0.66 times 0.34 all over our new sample size of 600 which this time we get a different answer of point 9614 and this is a higher probability, higher probability because our standard air is lower because again