Question

Question 6

A random sample of 516 U.S. businesses found that 144 of the 516 businesses had fired workers for misuse of the Internet. A 90% confidence interval for the proportion of all U.S. businesses who have fired workers for misuse of the Internet is between Blank-1 and Blank-2

Answer 1

Solution :

Given that,

n = 516

x = 144

Point estimate = sample proportion = = x / n = 144/516=0.279

1 -   = 1-0.279 =0.721

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

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

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

= 1.645 *((0.279*0.721) / 516)

= 0.0325

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.279- 0.0325< p <0.279+ 0.0325

0.2465< p < 0.3115

The 90% confidence interval for the population proportion p is : 0.2465, 0.3115