Question

A national survey of 1034 parents showed that 848 claimed pizza was their children's favorite food. Let p represent the proportion of all parents who claim that pizza is their children's favorite food. Use 2-decimal accuracy for all results.


The point estimate for p ( ) is ______.

The 98% confidence interval for p is (______ , ______).

Answer 1

Solution :

Given that,

n = 1034

x = 848

Point estimate = sample proportion = = x / n = 848/1034=0.820

1 -   = 1- 0.820 =0.18

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

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

= 2.326 (((0.82*0.18) / 1034)

= 0.03

A 98% confidence interval is ,

- E < p < + E

0.82-0.03< p < 0.82+0.03

0.79< p < 0.85

The point estimate for p ( ) is __0.82____.

The 98% confidence interval for p is (______ 0.79 , 0.85______).