Question

A 2011 poll found that 73​% of citizens of a certain country believe that high achieving high school students should be recruited to become teachers. This poll was based on a random sample of 1004 citizens. Complete parts a through d below.

a.) Find a​ 90% confidence interval for the proportion of citizens who would agree with this.

b.) Interpret your interval in this context.

c.) Explain what​ "90% confidence" means.

Answer 1

Solution :

Given that,

n = 1004

Point estimate = sample proportion = = 0.73

1 -   = 1- 0.73 =0.27

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 *((0.73*0.27) / 1004)

E = 0.0231

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.73- 0.0231< p < 0.73+0.0231

0.7069< p < 0.7531

The 90% confidence interval for the population proportion p is : 0.7069,0.7531