Question

Standard Normal (2) Distribution n POSITIVE z Scor 18 old 06 07 E- (Rou 5239 5636 5279 .5675 5438 5832 5871 6064 8968 .6406 6772 .6443 .6808 .7157 .7486 5065 7086 7123 1. 0 2 Standard Normal (2) Distribution: Cumulative Area from the LEFT 03 04 05 00 5000 5040 5080 5120 5160 5 199 5398 5478 5517 5557 5790 .5910 5948 5987 6179 6217 6255 6298 .6331 6564 6591 6628 6664 .6700 6736 6915 6960 7019 7054 .7257 .7291 7224 7357 749 .7580 .7611 .7642 7673 7704 7995 BS 8186 8212 1238 8254 8289 3413 3430 3461 345 8500 3531 3643 3565 3708 3569 8907 eges 1914 9049 3066 9062 9090 9115 207 9265 9332 9345 9357 9394 9452 9474 9484 94951 505 9564 3673 9591 9964 3719 9726 9738 9747 ST72 3778 9783 .9798 SEBEBSB.SE 7910 7867 .7454 7764 8061 8315 8554 .7794 8078 8340 8577 3729 3770 8790 9131 9147 9292 5236 8279 9406 18 9515 9525 3713 96.56 9750 Print Done Click to select your answers) Type here to search

Answer 1

we have given x= 526 , n = 955

a ) the best estimate of population proportion p is $\hat{p} =\frac{x}{n}$   

$\hat{p} =\frac{526}{955}$ = 0.551

b ) margin of Error (E) = $z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

from z table z at 90% confidence level is 1.64

$M.E. = 1.64\sqrt{\frac{0.551(1-0.551)}{955}}$

$M.E. = 0.026$

c ) the 90 % Ci for p = $\hat{p}\pm M.E.$

$= 0.551 \pm 0.026$

= (524,577)

90\%90% confidence interval for the population proportion is 0.524<p<0.577

d ) option B is true