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.
Construct a 90% confidence interval on the population proportion of correctly answered problems.
Construct a 95% confidence interval on the population proportion of correctly answered problems.
a) Solution :-
Given,
n = 100 ....... Sample size
= 0.78 ....... sample proportion.
Our aim is to construct 90% confidence interval.
c = 0.90
= 1- c = 1- 0.90 = 0.10
/2 = 0.10 2 = 0.05 and 1- /2 = 0.950
= 1.645 (use z table)
Now , the margin of error is given by
E = *
= 1.645 * [ 0.78 *(1 - 0.78)/ 100]
= 0.0681
Now the confidence interval is given by
( - E) ( + E)
( 0.78 - 0.0681 ) ( 0.78 + 0.0681 )
0.7119 0.8481
Required 90% Confidence Interval is ( 0.7119 , 0.8481 )
b) Solution :-
Given,
n = 100 ....... Sample size
= 0.78 ....... sample proportion.
Our aim is to construct 95% confidence interval.
c = 0.95
= 1- c = 1- 0.95 = 0.05
/2 = 0.025 and 1- /2 = 0.975
Search the probability 0.975 in the Z table and see corresponding z value
= 1.96
Now , the margin of error is given by
E = *
= 1.96 * [ 0.78 *(1 - 0.78)/ 100]
= 0.0812
Now the confidence interval is given by
( - E) ( + E)
( 0.78 - 0.0812 ) ( 0.78 + 0.0812 )
0.6988 0.8612
Required 95% Confidence Interval is ( 0.6988 , 0.8612 )
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 s