Question

A random sample found that forty percent of 100 Americans were satisfied with the gun control laws in 2017. Compute a 99% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2017. Fill in the blanks appropriately. A random sample found that forty percent of 100 Americans were satisfied with the gun control laws in 2017. Compute a 99% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2017. Fill in the blanks appropriately. A 99% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2017 is (____,____ ) (round to 3 decimal places)

Answer 1

Solution :

Given that,

n = 100

Point estimate = sample proportion = = 0.40

1 - = 1 -0.40 =0.60

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

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

= 2.576 (((0.40*0.60) / 100)

E = 0.126

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.40- 0.126< p < 0.40+ 0.126

0.274< p < 0.526

The 99% confidence interval for the population proportion p is : 0.274, 0.526