Question

your calculations. Answers may vary slightly due to rounding In a random sample of 68 professional actors, it was found that were extroverts (a) prepresent the proportion of all actors who are extroverts. Find testimates for ad d your tour de - 05735 ces (b) Find a 95% confidence interval for p. (Round your answers to two decimal places.) Find the maximal margin of error (Round your answer to two decimal places.) E. r Repe lower mit om the confidence interval for Round your answers to two decimal aces) Give abre interpretation of the meaning of the confidence interval you have found were confident that the true precortion of actor where extrovert a v We are confident that the true proportion of actors who are extroverts within the interval We are not that the proportionates where extrovert fall with this interval We are confident that the true proportion of actors who are extroverts a bove the interva (c) Do you think the conditions > Sand ng Sare satisfied in this problem? Fupiais why this would be an important considerat Yes, the conditions are satisfied S ertant tr i als us to say that is aromately mom Ne, the conditions are not satisfied. This is important because it was to say that approximately normal Yes, the conditions are said th a t becau s e in

Answer 1

Solution:

Given:

n = 68

x = 39

Part a) Sample proportion:

$\hat{p}=\frac{x}{n}$

P=

P=0.5735

then

q=1-p=1-0.5735 = 0.4265

ĝ= 0.4265

Part b) Find 95% confidence interval:

-E, P+E)

where

E = Margin of Error:

I êx E = Z XV PCV 12

We need to find z_c value for c=95% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.95) /2 = 1.95 / 2 = 0.9750

Look in z table for Area = 0.9750 or its closest area and find z value.

0.1 V123 .00 .01 .02 .03 .04 .05 .06 0.0 ,5000 15040 5080 5120 .5160 5199 39 1.5398 .5438 5478 1.5517 .5557 .5596 .636 0.2 15793 5832 ,5871 1.5910 .5948 1.5987 ,6026 0.3 .6179 .6217 .6255 .6293 .6331 1.6368 | 6406 0.4 .6554 ,6591 ,6628 .6664 .6700 .67366 16772 0.5 ,6915 ,6950 .6985 ,7019 .7054 7088 0.6 72577291 .7324 .7357 7389 7422 F454 0. 7 7580 7611 ,7642 .7673 704 7734 7764 0.8 1.788179107939 .7967 7995 ,8023 4051 09819816,8212 ,82388 .82648289 1.4315 18413 18438 ,8461 ,8485 .8508 18531 .4554 .8643 18665 8686 .8708 ,8729 ,8749 70 18849 1.8869 .888 ,8907 .8925 .894 962 1.3 19032 9049 ,9066 19082 19099 9115 1931 1.4 19192 .92079222 .9236 .9251 ,9265979 1.5 ,9332 19345 .9357 .9370 9382 19394 9606 1.6 19452 19463 19474 .9484 .9495 19505 15 1.7 ,9554 19564 19573 9582 19591 .9599 9508 1.8 19641 .9649 .9656 .9664 .9671 .96789 586 | 19 7 47- 0710 -07- 19750 2.0 ,9772 19778 19783 19788 19793 19798 19803 1.0 1.1 1.2

Area = 0.9750 corresponds to 1.9 and 0.06 , thus z critical value = 1.96

That is : Z_c = 1.96

Thus

I êx E = Z XV PCV 12

E = 1.96 X V 0.5735 x 0.4265 68

E = 1.96 X V0.00359696

E = 1.96 x 0.0599747

E = 0.11755

E= 0.12

Lower Limit =p-E

Lower Limit = 0.5735 - 0.12

Lower Limit = 0.4535

Lower Limit = 0.45

and

Upper Limit = p + E

Upper Limit = 0.5735+ 0.12

Upper Limit = 0.6935

Upper Limit = 0.69