Question

Suppose that insurance companies did a survey. They randomly surveyed 430 drivers and found that 330 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up.

a) x = ? b) n = ? c) p' = ? d)

Which distribution should you use for this problem? (Round your answer to four decimal places.)

P- (_) (_,_)

e) Construct a 95% confidence interval for the population proportion who claim they always buckle up.

i. state the confidence interval

ii. sketch graph

iii. calculate error bound

Answer 1

Solution:

Given:

Sample size =n = 430

x = Number of drivers always buckle up = 330

Part a) x = ?

x = 330

Part b) n = ?

n = 430

Part c) p' = ?

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

330 430

P=0.7674

Part d) Which distribution should you use for this problem?

Sampling distribution of sample proportions is approximate Normal distribution with mean of sample proportion:

Mộ=P

Mp= 0.7674

and standard deviation of sample proportions is:

px (1-P) Og = V!

0.7674 x (1 – 0.7674) Op = V 430

Op = 0.7674 X 0.2326 430

0 = V0.000415058

Op = 0.020373

Op = 0.0204

Thus

P ~ ( approximate Normal ) (
Mp= 0.7674
,
Op = 0.0204
)

Part e) Construct a 95% confidence interval for the population proportion who claim they always buckle up.

-E, P+E)

where

E = Zcx op

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

E = Zcx op

E = 1.96 x 0.020373

E = 0,0399

Thus

-E, P+E)

(0.7674-0.0399.0.7674 +0.0399)

(0.7275,0.8073)

i) a 95% confidence interval for the population proportion who claim they always buckle up is between

(0.7275,0.8073)

ii) sketch graph

95% confidence interval P-E = 0.7275 = 0.7674 + E = 0.8073

iii) calculate error bound

E = Zcx op

E = 1.96 x 0.020373

E = 0,0399