Question

[20 pts] In a random poll taken in 2008, Gallup asked 1010 national adults whether they were baseball fans. 570 of the sample said they were. Construct a 96% confidence interval to estimate the proportion of national adults who are baseball fans. Use 4 non-zero decimal places in your calculations.

a)Za/2

b)Find σp̂

c)Find the margin of error and construct the confidence interval

Answer 1

Solution :

Given that,

n = 1010

x = 570

Point estimate = sample proportion = = x / n = 570 / 1010 = 0.564

1 - = 1 - 0.564 = 0.436

a) At 96% confidence level

= 1 - 96%

=1 - 0.96 =0.04

/2 = 0.02

Z/2 = Z0.02 = 2.054

b)   [p ( 1 - p ) / n] = [(0.564 * 0.436) / 1010 ] = 0.0156

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

= 2.054 (((0.564 * 0.436) / 1010 )

= 0.0320

A 96% confidence interval for population proportion p is ,

± E

0.564 ± 0.0320

(0.532 , 0.596)