STAT 200 STAT200 Week 6 Homework

8.1.4 Suppose you compute a confidence interval with a sample size of 100. What will happen to the confidence interval if the sample size decreases to 80? Decreasing the sample size causes the error bound to increase, making the confidence interval wider. 8.2.6 In 2008, there were 507 children in Arizona out of 32,601 who were diagnosed with Autism Spectrum Disorder (ASD) ("Autism and developmental," 2008). Find the proportion of ASD in Arizona with a confidence level of 99%. X = 507 N = 32,601 (P+Z a/2 * sqrt p(1-p)/n) p = x/n p = 507/32,601 p = .016 a/2 = 1-.99/s = .005 Z.005 = +2.58 (.016 + 2.58 * sqrt .016(1-.016) / 32,601) = (.016 + .002) = (.014, .018) 8.3.2 Many people feel that cereal is healthier alternative for children over glazed donuts. Table #8.3.5 contains the amount of sugar in a sample of cereal that is geared towards children ("Healthy breakfast story," 2013). Estimate the mean amount of sugar in children cereal using a 95% confidence level. X + ta/2, n-1 * s/sqrt n n = 19 x = amount of sugar in cereal x = 212/19 = 11.158 s = sqrt E(x-x)^2/(n-1) = sqrt 156.527/(19-1) = 2.949 df = 19-1 = 18 Conf. Lvl = .95 a = (100-95)/100 = .05 a/2 = .025 t.025,18 = 2.101 x+t.025,18*s/sqrt n = 11.158+2.101*2.949/sqrt 19 = 11.158+2.101*.6765 = 11.158+1.421 = (9.737, 12.579) x x-x (x-x) ^2 10 – 11.158 = -1.158 1.341 14 – 11.158 = 2.842 8.077 12 – 11.158 = .842 .709 9 – 11.158 = -2.158 4.657 13 – 11.158 = 1.842 3.393 13 – 11.158 = 1.842 3.393 13 – 11.158 = 1.842 3.393 11 – 11.158 = -0.158 .025 12 – 11.158 = .842 .709 15 – 11.158 = 3.842 14.761 9 – 11.158 = -2.158 4.657 10 – 11.158 = -1.158 1.341 11 – 11.158 = -0.158 .025 3 – 11.158 = -8.158 66.553 6 – 11.158 = -5.158 26.605 12 – 11.158 = .842 .709 15 – 11.158 = 3.842 14.761 12 – 11.158 = .842 .709 12 – 11.158 = .842 .709STAT 200 Week 6 Homework Problems 9.1.7 A child dying from an accidental poisoning is a terrible incident. Is it more likely that a male child will get into poison than a female child? To find this out, data was collected that showed that out of 1830 children between the ages one and four who pass away from poisoning, 1031 were males and 799 were females (Flanagan, Rooney & Griffiths, 2005). Do the data show that there are more male children dying of poisoning than female children? Test at the 1% level. 1) x1 = number male children dying from accidental poisoning x2 = number of female children dying from accidental poisoning p1 = proportion of male children dying from accidental poisoning p2 = proportion of female children dying from accidental poisoning 2) Ho: p1 = p2 Ha: p1 / p2 a = .01 3) Assumptions a) random? YES b) binomial for each? NO c) 2 samples independent? YES d) at least 5 on each side for both? YES 4) Test Stat Sample Stat n1 = 1830 n2 = 1830 p1 = 1031/1830 = .563 p2 = 799/1830 = .437 q1 = 1-.563 = .437 q2 = 1-.437 = .563 pvalue p = x1+x2/n1+n2 = 1031+799/1830+1830 = 1830/3660 = .5 q = 1 - .5 = .5 z = (p1-p2)-(p1-p2)/sqrt pq/n1+pq/n2 = (.563-.437)-0 / sqrt . 5(.5)/1830+.5(.5)/1830 = .126 / sqrt .25/1830 + .25/1830 = .126 / .017 = 7.412 2(normalcdf(7.412, , 0, 1)) = 1.3E-13 1.3E-13