Question

Using the information from above, with p′=0.714, q′=0.286, and n=140, what is the 90% confidence interval for the proportion of the population who use public transportation?

z0.10	z0.05	z0.025	z0.01	z0.005
1.282	1.645	1.960	2.326	2.576

Use the table of common z-scores above.

• Round the final answer to three decimal places.
• 90% of the X-values lies between ____ and _____

Answer 1

Solution :

Given that,

n = 140

= 0.714

q′ = 0.286

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 * (((0.714 * 0.286) / 140) = 0.0628

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.714 - 0.0628 < p < 0.714 + 0.0628

0.6512 < p < 0.7768

The 90% confidence interval for the population proportion p is : 0.6512 and  0.7768