Answer:
Explanation:
Using the formula
p +/- (z* √[p(1-p) /n]
Where p is sample proportion = 120/400= 0.3 z*= 1.96 (z score for 95% confidence) and n is 400.
0.3 + (1.96 √[0.3(1-0.3) / 400])
0.3 + (1.96 √[(0.3*0.7)/400
0.3 + (1.96√ (0.21/400))
0.3 + (1.96 √0.000525)
0.3 + (1.96* 0.023)
0.3 + (0.045)
= 0.345 ~ 35%
For the lower interview
0.3 - (0.045)
= 0.255 ~ 26%
Thus, a 95% confidence interval for this study is between 26% and 35%