Question

Wcation manner for full credit; you must n an org s for credit. 1. Peanut allergies are considered to be some of the most serious food allergies National Institute of Allergy and Infectious Diseases, roughly 2% of Americans are According to the allergic to peanuts. many Americans must be sampled to ensure that the distribution of the sample proportion, p will be approximately normal? (3 pts) b) Suppose a sample of 550 Americans is obtained. For the distribution of the sample proportion, p, find the following, rounding to 4 decimal places as needed. (5 pts) Center Spread c) what is the probability that at least 396 of the 550 peopled surveyed had peanut allergies? (5 p ts) d) Suppose 7 of the 550 surveyed had peanut allergies. i) Would it be unusual for this many or even fewer of the 550 surveyed to have peanut allergies? Show how you arrived at your answer for credit. (3 pts) unusual. i) Explain why your answer to question (i) above is not necessarily evidence that the proportion Americans with peanut allergies has decreased? (2 pts)

Answer 1

$(a)~The~distribution~of~sample~proportion~will~be~approximately~normal~if\\ np\geq 10~or,~n\geq 10/p=10/0.02=500~where,~p=0.02.\\ At~least~500~American~must~be~sampled~to~ensure~that~the\\ ~distribution~of~sample~proportion~will~be~approximately~normal.\\ (b)~Since~550p=550*0.02=11>10~so\\ sampling~distribution~of~\hat{p}~is~Normal~distribution~with\\ Center=mean=p=0.02,\\~Spread=standard~deviation=\sqrt{\frac{p(1-p)}{550}}=0.0060.\\ (c)~P(\hat{p}\geq 0.03)=1-\Phi\left((0.03-0.02)/0.0060 \right )=0.0478.\\ (d)~(i)~\hat{p}=7/550=0.0127\\ Z-score=\frac{0.0127-0.02}{0.0060}=-1.2167\\ Since~Z-score~lies~between~(-3,3)~so~it~is~not~unusual.\\ (ii)~p-value=P(Z<-1.2167|Z\sim N(0,1))=0.1119>0.05\\ so~we~conclude~that~this~is~not~necessarily~evidence~that~the~proportion~of~Americans\\ with~peanut~allergies~has~decreased.$