Question

Question 8 1 pts To help consumers assess the risks they are taking the Food and Drug Administration (FDA) publishes the amount of nicotine found in all commercial brands of cigarettes. A new cigarette has recently been marketed. The FDA tests on this cigarette gave a mean nicotine content of 28.5 milligrams and standard deviation of 2.8 milligrams for a sample of n = 9 cigarettes. The FDA claims that the mean nicotine content exceeds 31.7 milligrams for this brand of cigarette, and their stated reliability is 95%. Do you agree? No, since the value 31.7 does not fall in the 95% confidence interval. Yes, since the value 31.7 does not fall in the 95% confidence interval Yes, since the value 31.7 does fall in the 95% confidence interval. No, since the value 31.7 does fall in the 95% confidence interval.

Answer 1

det Xi be a sample of cigarettes. i=1019 Then xin N( 2 g15, (2 gì?) The confidence interval is given ley. <uk X + Cd/2. In 6 6 ( * = lnd. ) X- ed/2 in 2.2 2. 2 = 1-2 2.8 28.5 - 1.645 X 3 syp < i < 28.5 +1.6458 - P ( 28.5-1.53 <ul < 28.5+.1.53)= 1-2 -i-d. , where a=0.05 ay p po (26197 3 (26.97 < Us 30.035 . MO, since the value 31.7 does not fall in the interval, How to find the confidence interval force. o2 is known dit Xw Nie, 02) X NNl4, 6%) 7 =l-d Now, Р (redbi L 6 pl P X V (T, 41) mle mele N N10,1), vn ( X -U) Cak) Calle <-4 < -* + PNE) = 1-4 p( x - I real & a/2 cili ext 2 Cd / ed/2) = 1-2 The confidence interveel for u given I need un i's by & red/21 ñt in CS Scanned with CamScanner