deontehiggins42
22.04.2020 •
Mathematics
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. Using the 95% level of confidence, what is the confidence interval for the proportion of students supporting the fee increase
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Ответ:
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' =
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
The Z-score at 95% level of significance =1.96
(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)
Ответ:
I got (0.5,1.5)
Step-by-step explanation:
I split the distance but I'm not quite sure I did it right.