Question

A large university is interested in the outcome of a course standardization process. They have taken a random sample of 100 student grades, across all instructors. The grades represent the proportion of problems answered correctly on a midterm exam. The sample proportion correct was calculated as 0.78.

a. Construct a 90% confidence interval on the population proportion of correctly answered problems.

b. Construct a 95% confidence interval on the population proportion of correctly answered problems.

Answer 1

Solution :

Given that,

n = 100

Point estimate = sample proportion = =0.78

1 -   = 1-0.78 =0.22

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 *((0.78*0.22) /100 )

E = 0.0681

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.78-0.0681 < p <0.78+ 0.0681

0.7119< p < 0.8481

The 90% confidence interval for the population proportion p is : 0.7119, 0.8481

(B)

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 *((0.78*0.22) /100 )

= 0.0812

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.78- 0.0812< p < 0.78+ 0.0812

0.6988< p <0.8612

The 95% confidence interval for the population proportion p is : 0.6988,0.8612