States. (a) A random sample of n 100 couples will be selected from this populati

Question

States. (a) A random sample of n 100 couples will be selected from this population and p, the proportion of couples that are mixed racially or ethnically, will be computed. What are the mean and standard deviation of the sampling distribution of p? (Round your standard deviation to four decimal places.) mean standard deviation (b) Is it reasonable to assume that the sampling distribution ofp is approximately normal for random samples of size n = 100? Explain. O Yes, because np 10 and n(1-p) > 1o. No, because np 10 No, because np > 10. (c) Suppose that the sample size is n-300 rather than n = 100, as in Part (b). Does the change in sample size change the mean and standard deviation of the sampling distribution of ? what are the values for the mean and standard deviation when n sOOP Round our standard deviation to four decimal places. mean standard deviation (d) Is it reasonable to assume that the sampling distribution of is approximately normal for random samples of size n = 300? Explain. O Yes, because np 10. No, because np 10 No, because np > 10 (e) When n-300, what is the probability that the proportion of couples in the sample who are racially or ethnically mixed wll be greater than 0.05? (Round your answer to four decimal places.)

Explanation / Answer

here p=0.04 , n=100 so

(a) mean=np=100*0.04=4 and

standard deviation=sqrt(npq)=sqrt(100*0.04*(1-0.04))=1.96

(b) right choice is third np<10

(c) mean=np=300*0.04=12 and

standard deviation=sqrt(npq)=sqrt(300*0.04*(1-0.04))=3.3941

(d) second np>10

(e) for p=0.05, the standard normal variate z=(P-p)/SE(p)

SE(p)=sqrt(p(1-p)/n)

z=(0.05-0.04)/sqrt(0.04*(1-0.04)/300)=0.8839

P(z>0.8839)=1-P(z<0.8839)=1-0.8116=0.1884

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