Question

We observed 280 successes in 700 independent trials. Compute a 90% confidence interval for the population proportion p.

Please answer in detail & explain the steps!

Answer 1

Solution :

Given that,

n = 700

x = 280

Point estimate = sample proportion = = x / n = 280/700=0.4

1 - = 1 - 0.4=0.6

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z _{0.05 = 1.645} ( Using z table ( see the 0.05 value in standard normal (z) table corresponding z value is 1.645 )

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

= 1.645 (((0.4*0.6) /700 )

E = 0.0305

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.4 - 0.0305 < p <0.4 + 0.0305

0.3695< p < 0.4305

The 90% confidence interval for the population proportion p is : lower limit =0.3695,upper limit =0.4305