In: Statistics and Probability
A sociologist sampled 200 people who work in computer related jobs, and found that 42 of them have changed jobs in the past 6 months. Construct a 85% confidence interval for the proportion of those who work in computer-related jobs who have changed jobs in the past six months
Solution :
Given that,
n = 200
x = 42
Point estimate = sample proportion = = x / n = 42/200=0.21
1 - = 1-0.21 =0.79
At 85% confidence level the z is ,
= 1 - 85% = 1 - 0.85 = 0.15
/ 2 = 0.075
Z/2 = Z0.075 = 1.44 ( Using z table )
Margin of error = E = Z/2 * ((( * (1 - )) / n)
= 1.44(((0.21*0.79) /200 )
E = 0.0415
A 85% confidence interval for population proportion p is ,
- E < p < + E
0.21-0.0415< p <0.21+ 0.0415
0.1685< p < 0.2515
The 85% confidence interval for the population proportion p is :0.1685, 0.2515