Question

1) According to the national center on health statistics (2004), 22.4% of adults are smokers. A random sample of 100 adults is obtained. State the criteria for normality and determine if it has been met in this case. (2pts) by Find the mean and standard error for the sampling distribution of sample proportions (2pts) In a random sample of 100 adults, what is the probability that at most 15 of them are smokers? (4pts) What is the probability of getting the same sample proportion from a sample of 500 adults? (Hint: you will need to recalculate the standard error) (6pts) 2) A Gallup poll in 2010 found that 493 of the 1050 Americans polled believed that the federal government is responsible for making healthcare available to all citizens. What is the point estimate for the proportion of Americans that believe the federal government is responsible for health care? (2pts) Determine the margin of error of your estimate AND write the confidence interval using interval notation. (4pts) State your conclusion in the context of the problem. (2pts)

Answer 1

Solution : Guen that population parameter p= 0.224 (ie 224%) Let x = Adult Smoker. H = p and Op = VPC-p. @) Criterla for normality is geen below. li) np > 10 (ii) n(1-P)210 to check normality (i) = 100 101 224) 210 22.4 310 (11 ) = 100(1-0.224170 770620 hence normality is proved (b) mean of Sample proportion , le Mp = ? Cukknow My = np and 6x = rmp (1-p HA = Mx 'ñ le Mp = mp = P ie, MP = 0.224. end Slendard ervor Of = ? 17. ñ op - 6x JP/IP hz = VAPID Of = 70.224 X 0.78 ( 0.776 100

000417 06 = 0.0417 C) eie want to find PCÔ40.15) = ? we know ito and of the correeflon factor = 6.050_ 0.50 100 le = 0.005 adjusted Right limita 0.15+0.005 + 0.155. P ( Å < 0.15) P.($20.155) → Þ($-0.2.44 < 0.155-0244) 0.0417 = P(Z = -1065) PCB 40.15) = 0.049. d). p (<0.15) = ? for n=900. Mi = 0.2244 66 - (1-P) - 10.24 (0.146) V 500 Op = 0,0187 cf = 050 = 0.001 adjusted Right Imil - 0.15+0.001 = 0.151 P (2015) n P ($50.151) =p ( B-0.224 < 0.151 - 000244 ) 0.0187 0.0187

an = P(Ze-3.92) P(P<0.05) = 0