Question

900 college freshmen were randomly selected for a national survey. Among survey participants, the mean grade-point average (GPA) was 3.3, and the margin of error was +4% for the 80% confidence interval. If we want to get a 90% confidence interval with a small margin of error t4%. How many students do you need to select?

Answer 1

Solution:

Given: for 80% confidence level , margin of error = E = +/- 4% and sample size = n = 900

We have to find sample size n for 90% confidence interval with same margin of error +/-4%.

$n= \left ( \frac{Z_{c}\times \sigma }{E} \right )^{2}$

Population standard deviation is not given but we can find it from given information.

Use Margin of Error formula:

E = ếc X 3Vn

For c = 80% , find area = ( 1 + c) / 2 = ( 1+0.80) / 2 = 1.80 /2 = 0.90

Look in z table for area = 0.9000 or its closest area and find z value:

.00 .01 07 | .08 .09 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 15000 .5040 .5398 .5438 .5793 .5832 .6179 .6217 .6554 6591 .6915 .6950 72577291 .7580 7611 7881 .8159 ,8186 .8413 8438 .8643 ,8665 19840 8860 9032 ,9049 .5080 .5120 .5478 .5517 ,5871 .5910 .6255 .6293 .6628 .6664 .6985 ,7019 .7324 .7357 ,7642 .7673 1.7967 .8212 .8238 ,8461 .8485 .8686 .8708 8888 -9997 9066 9 082 .5160 .5199 .5239 5279 .5557 .5596 5636 .5675 .5948 .5987 .6026 .6064 .6331 .6368 .6406 .643 .6700 .6736 ,6772 .6808 .7054 7088 ,7123 7157 .7389 7422 .7454 7486 74 734 ,764 .7794 .7995 .8023 .8051 .8078 .8264 18289 ,8315 ,8340 ,8508 .8531 .8554 .8577 ,8729 .8749 ,8770 .8790 -6955894.8962.6980- .9099 9115 9131 9147 .5819 .5*14 .6103 .6180 .684 7190 $17 7823 .8106 .8365 .8999 18810 1.8997) .9162 15359 5753 .6141 1,6517 .6879 ,7224 .7549 7852 .8133 .8389 ,8621 1.8830 19015 ,9177

Area 0.8997 is closest to 0.9000 and it corresponds to 1.2 and 0.08

thus z_c = 1.28

Thus

E = ếc X 3Vn

0.04 = 1.28 xσ/V900

0.04 = 1.28 xσ/30

0.04 x 30/1.28 = 0

0,9375 = σ

that is:

O = 0.9375

Now find sample size for 90% confidence level;

$n= \left ( \frac{Z_{c}\times \sigma }{E} \right )^{2}$

Z_c is z critical value for c = 90% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.90) / 2 = 1.90 / 2 = 0.9500

Look in z table for Area = 0.9500 or its closest area and find corresponding z value.

. | 0.0 10.1 10.2 10.3 0.4 | 0.5 0.6 0.7 0.8 0.9 .01 .5040 5438 .5832 .6217 6591 6950 .02 .5080 5478 .5871 .6255 .6628 .6985 | .03 .5120 .5517 .5910 .6293 6664 .7019 .04 5160 .5557 .5948 6731 | 1.6100 .7054 .00 .5000 .5398 | 5793 1.6179 6554 6915 7257 .7580 7881 .8159 1.8413 ,8643 .8849 9032 .9192 1.9332 9452 9554 .05 .06 .5199 .5239 .5496 5636 5987 .6026 .6468 6406 636 .6772 .7038 | 1.7123 1.7454 .7734 .7764 .8023 1.8051 8051 | .8289 | 1.8315 .8551 .8554 .8749 .8770 .8944 .8962 .9115 .9131 .9255 .9279 .9394 .9406 9505 .9515 9599 9608 .07 .5279 .5675 6064 6443 .6808 7157 7486 .7794 .8078 .8340 .8577 .8790 .8980 9147 .9292 .9418 .9525 .9616 7611 1.7910 .8186 .8438 ,8665 .8869 .9049 9207 9345 9463 .9564 10 .08 .09 .5319 .5359 .5714 .5753 .6103 .6141 6480 .6517 .6844 .6879 .7190 .7224 .7517 .7549 .7823 78521 8106 .8133 .8365 .8389 .8599 .8621 8810 8830 .8997 19015 .9162 | 19177 9306 | 9319 .9429 | 19441 9535 .9545 9625 9633 .7642 .7673 704 .7939 .7967 | 17295 .8212 .8238 .8264 .8461 1.8485 .8508 8686 .8708 .8729 .8888 | .8907 .8425 .9066 .9082 .9099 .9222 .9236 9251 9357 9370 .9382 9474 19484 | ..9495 .9573 .9582 .9591 8412 1.1 1.2 1.3 1.4 1.5 1.6 1.7

Area 0.9500 is in between 0.9495 and 0.9505 and both the area are at same distance from 0.9500

Thus we look for both area and find both z values

Thus Area 0.9495 corresponds to 1.64 and 0.9505 corresponds to 1.65

Thus average of both z values is : ( 1.64+1.65) / 2 = 1.645

Thus Z_c = 1.645

Thus

$n= \left ( \frac{Z_{c}\times \sigma }{E} \right )^{2}$

(1.645 x 0.9375 0.04

n = (38.55469

n = 1486.463928

n = 1487

Thus we need to select 1487 students.