Question

The Arizona Department of Transportation (ADOT) is considering changing statewide speed limits from 65 mph to 75 mph. They wish to perform analysis on p, the proportion of Arizona residents that favor this change. In order to prove that a majority of residents favor the change, ADOT wishes to test the following hypotheses using a level of significance of 0.05. Ho:p = 0.5 H:p > 0.5 ADOT wishes to design the test to have a power of 90% if the true population proportion is equal to 0.55. A) How many residents should be surveyed? You may use this sample size for the remainder of this problem. (Note: You should get a value between 800 and 900 on this part. If you did not, then choose a value between 800 and 900 and use it to work the remainder of the problem.) B) Suppose the sample size found in part a is implemented and 461 of the surveyed residents favor the change. Construct a point estimate on the proportion of interest. C) Identify the test statistic and critical region for the test. State the conclusion for this test. D) What is the P-value for this test? State the conclusion for this test. E) Construct an appropriate confidence interval to verify your conclusion from part c. F) We wish to use a 95% lower confidence interval to estimate p. How many residents should be surveyed so that the error in our estimate is less than 4%.

Answer 1

change. At beet the the change coe not favoring favoring and Toe the above bet be the population proportion of Arizona Residents that Aesume, n is the total number of surveyed person which large. given level of significance. 4-0-05, me like to test- Ho : p=0.5 against #i: Þ>ou5 to find the test statistic, we know, Ŝ which is noth othing Sample proportion estimate of Arizona Residents that favour it's obvious E(E)= p and voor (&) = $(1-6) since efo define a dieho tomoues Bernoulli variable for women get since se is large, by CLT, Z. - E(6) (out) Var (6) ie Valo -6) Za(0, 1) render Ho, if he is large. √(1-P) and under Ho, Z can be resed as test statistic. critical region. We rejeet Ho if observed zye at level X=0.05, where eis 3, Pito (z>e). «=0:05 , 7 (0,1) e-To.05 = 1.645 - upper.x. point of N(0,1). therefore, we reject Ho, if Êx Pro+ 1.645. P (1- Filo) but PHO=0.5, henee reject to at *20.05, if Ê > 0-5+ 0-8225 otherwise aecept it 1 a). given toue population proportion. p =0-55 Power - PH ( reject. Ho). $ 0.5 + E-p > 0.5 +0.8225 0.8225 PHI p Pue Feb n 1(77(0-5-0-55) + 0.8225 0.55 X0.45 10.55x0.45 TL ZNN(0,1) and PH 0.55

i power P(ZY 1.6533 -0.1005 Jn $ (0-1005 Jn - 106533 which is to 0.9. m2 = 857. FO30 - 858 P = 461 461 858 = 0.5373. n (F(x) + F(-2)=) 0-1005 in) 1 - $ (106533 - 0-100518) given equal (147) (0-1005 -16533) - 0-9 0-10051h - 1.6533 = 129 (from table). (ARS.) (6) point estimate on the proportion of interest (Ans.) les .We have alreadey developed the test statistie as N(o, i no Jp (1-6) evitical region given by.c), √ 858 { : 8 x 0.5281} on the basis of paot (b), ß = 0·5373 70.5281, dejeet Ho at 2=0.05 and believe that population poopoštion is greater than Ya. (d) p-value Pita (Z* > observed z). 7*».060,1). render Ho, observed z. 1858 (0.5373-005) 10.500.5 Z-V5 (66) Ho W YS: : € > 0.5 +0.8225 } i.e conelusion: we = 2.1852. Pius (7> 2• 1852) - 1- PH. (2* <2-1852) 1 - (2-1852) 21-0.9857 = 0.0143 <x=0.05 Conelusion. Sinee p-value < 420.05, reject Ho at & = 0.05 we -