In: Statistics and Probability
Exercise 2. A survey of 525 homeless persons showed that 168 were not veterans. Construct a 95% confidence interval for the proportion of homeless persons who are not veterans.
Solution :
Given that,
n = 525
x = 168
Point estimate = sample proportion = = x / n = 0.32
1 - = 0.68
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 * (((0.32*0.68) / 525)
= 0.040
A 95% confidence interval for population proportion p is ,
- E < p < + E
0.32 - 0.040 < p < 0.32 + 0.040
0.280 < p < 0.360
The 95% confidence interval for the population proportion p is : ( 0.280 , 0.360 )