Question

In a simple random sample of 70 automobile dealers registered in a certain state, 30 of them were found to have emission levels that exceeded a state standard. At the 95% confidence interval, the proportion of automobiles whose emissions exceeded the state standard is closest to?

Answer 1

Solution :

Given that,

n = 70

x = 30

Point estimate = sample proportion = = x / n = 30/70=0.429

1 -   = 1- 0.429=0.571  

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 (((0.429*0.571) / 70)

E = 0.116

A 95% confidence interval is ,

- E < p < + E

0.429-0.116 < p < 0.429+0.116

(0.313 , 0.545)