Question

4. The National Center for Health Statistics interviewed 5409 adult smokers in 2015, and 2636 of them said they had tried to quit smoking during the past year. Consider this to be a random sample Find a 95% confidence interval for the proportion of smokers who have tried to quit within the past year. Find a 99% confidence interval for the proportion of smokers who have tried to quit within the past year а. b. Someone claims that the proportion is less than 0.5. With what level of confidence can this statement be made? с. d Find the sample size necessary for a 95% confidence interval to specify the proportion to within +0.01 Find the sample size necessary for a 99% confidence interval to specify the proportion to within ±0.01 е.

Answer 1

n=5409 X=2636 Z-table 0.49 X0.51 5409 4 Given that : n=5409 P = X = 2636 = 10.49 n - 5409 @ The 95% confidance interval for 'p' (+24, JPCC) ] + [ 0.49 + (-26 ) forugko-ST! 24, wing 2 table? + (0.49 + 1-46 X 0.0067977 (0.49 ± 0.0133 [[ 0.4767, 0.5033] 6 The 99% confidance interval for 'p' [ Ể + Zube J Pline? * (0:49 (2-545) 104pxos 1 ] {24, wing z-table? *(0.49 † (2.575) (0.006797)] [0.49 ± 0.01750 ] ] + [ 0.4725, 0.5075] 6 90% contidance level this statement made

@ Given that - margin of enor (E) = groot 0.01 d =0.05 Î = 0.49 we know, E = Zadly n s za wing etable? 0.01 = (1.96) √ 0.49X0.51 Vn 0.01 © n = (-1 ) ( - 4331) Vn : 0.91 010) Tn = 97.98 na 9600 Given that :- margin of emor( e ) = 0.01 2=0.0 À = 0.49 'E= 2x1₂ SPEED 0.01- (2575) (0.4999) we know, 7 PCI-P) 1.2872 in= 3.374 - 128.72 0.01 nu 16569.9 Inu 18570