Question

Use the sample data and confidence level given below to complete parts​ (a) through​ (d)

A research institute poll asked respondents if they felt vulnerable to identity theft. In the​ poll, n=1056 and x=599 who said​ "yes." Use a 90% confidence level.Use the sample data and confidence level given below to complete parts​ (a) through​ (d)

Answer 1

Solution:

Given:

n = 1056

x = 599

Confidence level = c = 90%

Part a) Find point estimate of population proportion:

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

$\hat{p}=\frac{599}{1056}$

P = 0.567

Part b) Margin of error:

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

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.

.00 101 .02 103 106 107 108 209 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 13 1.4 1.5 1.6 1.7 .5000 1.5398 .5793 .6179 .6554 .6915 7257 47580 7881 18159 .413 18643 18849 19032 19192 19332 CACO 19554 15040 .5438 1.5832 .6217 .6591 .6950 7291 7611 7910 1.8186 .8438 .8665 .8869 19049 19207 19345 0463 19564 15080 15478 .5871 .6255 .6628 .6985 .7324 7642 7939 1.8212 1.8461 1.8686 18888 1.9066 1.9222 19357 19474 1973 1.5120 .5517 .5910 .6293 .6664 17019 7357 .7673 .7967 18238 .8485 8708 18907 1.9082 1.9236 19370 9484 19582 1.04 105 sl60 5199 1.5557 5996 5948 1.5987 .6331 6368 .600 .6766 7054 7088 17889 7412 .T04 734 7095 .80% .8264 18288 .8508 .851 .8/29 18749 .8925 1.894 1.9099 .9115 1951 1.925 19382 1934 19495 19505 19591 19599 15239 1.5636 .6026 .6406 1.6772 7123 7454 7764 .8051 .8315 18554 .8770 .8962 19131 1.9279 19406 19515 19608 1.5279 1.5675 .6064 .6443 1.6808 7157 7486 7794 .8078 1.8340 .8577 1.8790 1.8980 19147 1.9292 19418 9525 19616 1.5319 5714 1.6103 .6480 .6844 1.7190 7517 47823 18106 .8365 1.8599 1.810 .8997 .9162 19306 19429 19535 19625 15359 5753 .6141 .6517 .6879 .7224 7549 7852 18133 .8389 ,8621 1.8830 19015 19177 19319 1941 19545 19633

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

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

$E = 1.645 \times \sqrt{\frac{0.567 \times (1-0.567 )}{1056}}$

$E = 1.645 \times \sqrt{\frac{0.567 \times 0.433 }{1056}}$

$E = 1.645 \times \sqrt{0.000232462 }$

$E = 1.645 \times 0.0152467$

$\mathbf{{\color{Magenta} E = 0.0251}}$

Part c) Construct the confidence interval.

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

$\mathbf{{\color{DarkOrange} (0.542 \: \: < p < \: \:0.592 )}}$

Part d) Interpretation:

One has 90​% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.