Please help I’m so confused 25 points and brainliest!!
1 answer:
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Answer:
The confidence interval for the proportion of students supporting the fee increase
( 0.77024, 0.81776)
Step-by-step explanation:
<u>Explanation:</u>
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
<u>The 95% level of confidence intervals</u>
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)
<u>Conclusion:</u>-
The confidence interval for the proportion of students supporting the fee increase
( 0.77024, 0.81776)
