In: Statistics and Probability
A teacher wanted to estimate the proportion of students who take notes in her class. She used data from a random sample of size n = 76 and found that 45 of them took notes. The 97% confidence interval for the proportion of student that take notes is:
Solution :
Given that,
n = 76
x = 45
Point estimate = sample proportion = = x / n = 45 /76=0.469
1 - = 1-0.469=0.531
At 97% confidence level
= 1 - 97%
= 1 - 0.97 =0.03
/2
= 0.015
Z/2
= Z0.015 =2.170( Using z table )
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 2.170 (((0.469*0.531) /76 )
E = 0.1242
A 97% confidence interval for population proportion p is ,
- E < p < + E
0.469 - 0.1242< p < 0.469 +0.1242
0.3448< p < 0.5932
The 97% confidence interval for the population proportion p is : 0.3448, 0.5932