In: Statistics and Probability
Out of 100 people sampled, 14 had kids. Based on this, construct
a 90% confidence interval for the true population proportion of
people with kids.
Give your answers as decimals, to three places
< p <
Solution :
Given that,
n = 100
x = 14
Point estimate = sample proportion = = x / n = 14/100=0.14
1 - = 0.86
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 *((0.14*0.86) /100 )
E = 0.057
A 90% confidence interval for population proportion p is ,
- E < p < + E
0.14-0.057 < p < 0.14+0.057
0.083< p < 0.197