Question

According to a study, 1156 out of 7730 adults said that they would ask their siblings first in case they need an emergency loan. Using these data, construct a 90% confidence interval. Also find the margin of error. Round your answers to 3 decimal places.

Группа выборов ответов

(14.3%,15.6%) margin of error=0.08%

(14.2%,15.7%), margin of error=0.8%

(14.2%,15.7%), margin of error=1.6%

(14.2%,15.7%), margin of error=16%

(14.3%,15.6%) margin of error=0.7%

Answer 1

Solution :

Given that,

n = 7730

x = 1156

Point estimate = sample proportion = = x / n = 1156/7730=0.150

1 -   = 1- 0.150 =0.850

At 90% confidence level the z is

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= 1.645 *((0.150*0.850) /7730 )

E = 0.7%

A 90% confidence interval for population proportion p is ,

- E < p < + E

15.0%-0.7% < p < 15.0%+0.7%

14.3%< p < 15.6%

The 90% confidence interval for the population proportion p is : 14.3%< p < 15.6%