Question

An interactive poll found that 318 of 2,308 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 98% 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 = 2308

x = 318

(a)

= x / n = 318 / 2308 = 0.138

1 - = 1 - 0.138 = 0.862

(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 * (((0.138 * 0.862) / 2308)

= 0.012

A 90% confidence interval for population proportion p is ,

- E < P < + E

0.138 - 0.012 < p < 0.138 + 0.012

0.126 < p < 0.150

(0.126 , 0.150)

(c)

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z_0.01 = 2.326

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

= 2.326 * (((0.138 * 0.862) / 2308)

= 0.017

A 98% confidence interval for population proportion p is ,

- E < P < + E

0.138 - 0.017 < p < 0.138 + 0.017

0.121 < p < 0.155

(0.121 , 0.155)

(d)

If we increase level of confidence the width of the confidence is increases .