Respuesta :
Answer:
The confidence interval for the proportion of students supporting the fee increase
( 0.77024, 0.81776)
Step-by-step explanation:
Explanation:
Given data a survey of an urban university (population of 25,450) showed that 883 of 1,112 students sampled supported a fee increase to fund improvements to the student recreation center.
Given sample size 'n' = 1112
Sample proportion 'p' = [tex]\frac{883}{1112} = 0.7940[/tex]
q = 1 - p = 1- 0.7940 = 0.206
The 95% level of confidence intervals
The confidence interval for the proportion of students supporting the fee increase
[tex](p-z_{\alpha } \sqrt{\frac{pq}{n} } ,p + z_{\alpha } \sqrt{\frac{pq}{n} } )[/tex]
The Z-score at 95% level of significance =1.96
[tex](0.7940-1.96\sqrt{\frac{0.7940 X 0.206}{1112} } ,0.7940 + 1.96 \sqrt{\frac{0.7940 X 0.206}{1112} } )[/tex]
(0.7940-0.02376 , 0.7940+0.02376)
( 0.77024, 0.81776)
Conclusion:-
The confidence interval for the proportion of students supporting the fee increase
( 0.77024, 0.81776)
