Question

8. A college wishes to know the proportion of American adults who speak two or more languages. The survey includes asking 565 randomly selected American adults and finds that 226 speak two or more languages. a) Find the point estimate of the percentage of adults who speak two or more languages. b) Find a 96% confidence interval estimate of the percentage of Americans who speak two or more languages. Does the majority of Americans speak two or more languages? Explain.

Please show hand written work for everything including formulas etc.

Answer 1

Solution:
Given: n = sample size of American adults = 565
x = Number of adults speak two or more languages = 226
Part a) Find Point estimate of the percentage of adults who speak two or more languages.

p =

226 р- 565

$\hat{p}=0.4$

Part b) Find 96% confidence interval estimate for the percentage of adults who speak two or more languages.

$( \hat{p} - E \: \: ,\: \: \hat{p} + E )$

where

$E = Z_{c}\times \sqrt{\frac{\hat{p}\times (1-\hat{p})}{n}}$

We need to find z_c value for c=96% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.96) /2 = 1.96 / 2 = 0.9800

Look in z table for Area = 0.9800 or its closest area and find z value.

.00 .01 .02 .03 .04 .05 .06 .07 .08 .09 Z 5040 .5080 5120 .5517 .519 .5596 5987 .6368 .6786 7088 7422 77 4 .8013 .8289 .8531 87-9 .8914 .9115 .9165 .9894 9505 9599 678 .9744 0.0 5000 .5160 .5239 5279 5319 .5359 .5478 5557 5636 5675 5714 .5753 0.1 .5398 5438 0.2 5793 5832 .5871 .5910 5948 .6026 6064 .6103 6141 6480 .6179 6217 .6255 6293 .6331 .6406 .6443 6517 .6879 0.3 6664 6808 6844 0.4 6554 6591 6950 .6628 .6985 .6700 .6772 7190 7517 7019 0.5 .6915 7054 7123 7157 7224 7454 7357 7486 7549 0.6 7257 7291 7324 7389 7704 7852 0.7 7580 7611 7642 7673 7764 7794 7823 8051 8315 8078 8106 7881 7967 7995 8133 0.8 7910 7939 .8264 8389 0.9 8159 8186 .8212 8238 8340 .8365 8577 .8413 .8438 .8461 8485 .8508 .8554 .8599 8621 1.0 .8810 .8830 1.1 8643 .8665 .8686 .8708 .8729 8770 8790 1.2 8907 8849 .8869 .8888 .8925 .8962 8980 .8997 .9015 .9066 .9131 .9147 .9162 9177 1.3 .9032 .9049 .9082 .9099 9292 9306 1.4 9192 .9207 .9222 .9236 .9251 .9279 9319 .9406 .9515 9441 .9418 .9429 1.5 9332 .9345 .9357 .9370 .9382 .9463 1.6 .9452 .9474 .9484 9495 .9525 .9535 .9545 .9573 .9582 .9591 .9608 .9616 .9625 .9633 1.7 9554 .9564 .9656 .9664 1.8 .9641 9649 .9671 .9686 .9693 .9699 .9706 .9750 .9767 .9713 .9719 .9726 .9732 .9738 .9756 .9761 1.9 9812 .9808 .9817 2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803

Area = 0.9798 is closest to 0.9800 and it corresponds to 2.0 and 0.05 , thus z critical value = 2.05

That is : Z_c = 2.05

Thus.

$E = Z_{c}\times \sqrt{\frac{\hat{p}\times (1-\hat{p})}{n}}$

$E = 2.05 \times \sqrt{\frac{0.4 \times (1-0.4)}{565}}$

$E = 2.05 \times \sqrt{\frac{0.4 \times 0.6}{565}}$

$E = 2.05 \times \sqrt{ 0.00042478 }$

$E = 2.05 \times 0.02061016$

Thus

$( \hat{p} - E \: \: ,\: \: \hat{p} + E )$

Thus a 96% confidence interval estimate for the percentage of adults who speak two or more languages is between :

Since both the limits are less 50% , there Majority of American adults do not speak two or more languages.