Question

in a random poll taken in 2008, Gallup asked 1006 national adults whether they were baseball fans. 433 of the sample said they were. Construct a 95% confidence interval​ to estimate the ​proportion​ of national adults who are baseball fans. ​Use 4 non-zero decimal places on calculations.

a) Find the Zα/2

b) Find σpˆ

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

Answer 1

Solution :

Given that,

n = 1006

x = 433

Point estimate = sample proportion = = x / n = 433/1006=0.430

1 -   = 1- 0.430 =0.570

At 95% confidence level

  = 1 - 95% = 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)

= 1.96 (((0.430*0.570) /1006 )

E = 0.0306

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.430-0.0306 < p < 0.430+0.0306

0.3994< p < 0.4606

The 95% confidence interval for the population proportion p is : 0.3994, 0.4606

standard error =σpˆ=(((0.430*0.570) /1006 )=0.0156