Question

A North Carolina university wants to determine how many of its graduating students go on to find gainful employment in their field of study. After sampling 200 students from the previous graduating class, they find that 174 of them are employed in a related field. Using this information, construct the 90% confidence interval for the amount of students that will go on to be employed in their field.

Answer 1

Solution :

Given that,

n = 200

x = 174

Point estimate = sample proportion = = x / n = 174/200=0.87

1 - = 1 -0.87=0.13

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645} ( Using z table )

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

= 1.645 (((0.87*0.13) / 200)

E = 0.0391

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.87- 0.0391< p < 0.87+0.0391

0.8309< p < 0.9091

The 90% confidence interval for the population proportion p is :0.8309, 0.9091