Question

A random sample of 900 college students found that 62% study outside of the class less than 2 hours per week. Construct a 90% confidence interval for the proportion of students who study less than 2 hours per week [5 points] . Interpret the results using complete sentences. [5 points]

Answer 1

Solution :

Given that,

n = 900

Point estimate = sample proportion = = 0.62

1 -   = 1- 0.62 =0.38

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 *((0.62*0.32) / 900)

E = 0.024

A 90% confidence interval for the proportion of students who study less than 2 hours per week

- E < p < + E

0.62 - 0.024 < p < 0.62+0.024

0.596< p < 0.644