According to a recent survey, 72% of teens ages 12-17 in a certain country used

Question

According to a recent survey, 72% of teens ages 12-17 in a certain country used social networks in 2009. A random sample of 110 teenagers from this age group was selected. Complete parts a through d below. a. Calculate the standard error of the proportion. Sigma_p = (Round to four decimal places as needed.) b. What is the probability that less than 74% of the teens from this sample used social networks? P(Less than 74% of the teens from this sample used social networks) = (Round to four decimal places as needed.) c. What is the probability that between 70% and 80% of the teens from this sample used social networks? P(Between 70% and 80% of the teens from this sample used social networks) = (Round to four decimal places as needed.) d. What impact would changing the sample size to 200 teens have on the results of parts a, b, and c? Choose the correct answer below. A. The standard error would be increased, which would, in turn, increase the probabilities that the sample proportions will be closer to the population proportion. B. The standard error would be increased, which would, in turn, reduce the probabilities that the sample proportions will be closer to the population proportion. C. The standard error would be reduced, which would, in turn, reduce the probabilities that the sample proportions will be closer to the population proportion. D. The standard error would be reduced, which would, in turn, increase the probabilities that the sample proportions will be closer to the population proportion. E. Changing the sample size would have no effect on the standard error or the probabilities that the sample proportions will be closer to the population proportion.

Explanation / Answer

Here p=0.72 and n=110

a. standard error=sqrt(p(1-p)/n)=0.043

b. Here we need to fin P(x<0.74)

We will convert x to z

P(z<0.74-p/standard error)=P(z<0.74-0.72/0.043)=P(z<0.47)=0.5-P(0.47<z<0)=0.5+0.1808=0.6808

c. P(0.70<x<0.80)=P(0.70-0.72/0.043<z<0.80-0.72/0.043)=P(-0.47<z<1.86)=P(0<z<1.86)-P(0<z<-0.47)=0.4686+0.1808=0.6494

d. Answer is d. The standard error would be reduced, which in turn, increase the probabilities that the sample proportions will be closer to the population proportion.

As n in SE is a denominator so big the number less the SE and further in finding probability again SE in denominator so as it reduces the probability reduces.

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