Question

An interactive poll found that 393 of 2,261 adults aged 18 or older have at least one tattoo. ?(a) Obtain a point estimate for the proportion of adults who have at least one tattoo. ?(b) Construct a 90?% confidence interval for the proportion of adults with at least one tattoo. ?(c) Construct a 99?% confidence interval for the proportion of adults with at least one tattoo. ?(d) What is the effect of increasing the level of confidence on the width of the? interval?

Answer 1

Solution :

Given that,

n = 2261

x = 393

a ) = x / n = 393 / 2261 = 0.174

1 - = 1 - 0.174 = 0.826

b ) At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 * (((0.174 * 0.826) / 2261) = 0.013

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.174 - 0.013 < p < 0.174 + 0.013

0.161 < p < 0.187

c ) At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.576 * (((0.174 * 0.826) / 2261) = 0.021

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.174- 0.021 < p < 0.174 + 0.021

0.153 < p < 0.195

d ) The increasing the level of confidence on the increasing width of the? interval.