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.

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

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

Answer 1

a) Solution :-   

Given,

n = 100 ....... Sample size

   $\hat p$ = 0.78 ....... sample proportion.

Our aim is to construct 90% confidence interval.

$\therefore$ c = 0.90

$\therefore$$\alpha$ = 1- c = 1- 0.90 = 0.10

$\therefore$  $\alpha$/2 = 0.10 $\slash$ 2 = 0.05 and 1- $\alpha$ /2 = 0.950

$\therefore$ = 1.645 (use z table)

Now , the margin of error is given by

E =  *  

= 1.645 * $\sqrt{}$ [ 0.78 *(1 - 0.78)/ 100]

= 0.0681

Now the confidence interval is given by

($\hat p$ - E)   ($\hat p$ + 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

   $\hat p$ = 0.78 ....... sample proportion.

Our aim is to construct 95% confidence interval.

$\therefore$ c = 0.95

$\therefore$$\alpha$ = 1- c = 1- 0.95 = 0.05

$\therefore$  $\alpha$/2 = 0.025 and 1- $\alpha$ /2 = 0.975

Search the probability 0.975 in the Z table and see corresponding z value

$\therefore$ = 1.96

Now , the margin of error is given by

E =  *  

= 1.96 * $\sqrt{}$ [ 0.78 *(1 - 0.78)/ 100]

= 0.0812

Now the confidence interval is given by

($\hat p$ - E)   ($\hat p$ + E)

( 0.78 - 0.0812 )    ( 0.78 + 0.0812 )

0.6988   0.8612  

Required 95% Confidence Interval is ( 0.6988 , 0.8612 )