In: Math
A random sample of 200 cars has 18 that are green in color. What is the confidence interval for the true proportion of green cars that lie within a 95% confidence interval?
Solution :
Given that,
n = 200
x = 18
Point estimate = sample proportion = = x / n = 0.09
1 - = 0.91
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
Z/2 = Z0.025 = 1.96
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.96 * (((0.09 * 0.91) / 200)
= 0.040
A 95% confidence interval for population proportion p is ,
- E < p < + E
0.09 - 0.040 < p < 0.09 + 0.040
0.050 < p < 0.130
The 95% confidence interval is ( 0.050 , 0.130 )