Answer :
The confidence interval for the proportion of students supporting the fee increase: (0.5827, 0.6539)
What is Interval?
In mathematics, an interval is expressed in numerical terms. All the numbers between two specific integers are referred to as an interval. All actual values between those two are included in this range. Any form of number you may imagine is a real number.
We have the following confidence interval of proportions for a sample of n people who were surveyed with a probability of success of π and a confidence interval 1 - α
π ± z[tex]\sqrt{\pi (1 - \pi )/n}[/tex]
, where
Z stands for the z score with a p value of 1 - α/2
We have this in relation to the issue:
A price increase to pay for upgrades to the student recreation center was supported by 780 of the 1,280 students who were sampled. So
n = 1280, π = 780 / 1280
π = 0.6093
95 percent confidence level
Consequently α = 0.05, z is the Z value that has a p value of, thus
1 - 0.05/2 = 0.975
So, z = 1.96
The interval's lowest limit is:
= π - z[tex]\sqrt{\pi (1 - \pi )/n}[/tex]
= 0.6093 - 1.96[tex]\sqrt{0.6093 * 0.3907/1280}[/tex]
= 0.6093 - 1.96[tex]\sqrt{0.000185}[/tex]
= 0.5827
This range's maximum value is:
= π + z[tex]\sqrt{\pi (1 - \pi )/n}[/tex]
= 0.6093 + 1.96[tex]\sqrt{0.6093 * 0.3907/1280}[/tex]
= 0.6093 + 1.96[tex]\sqrt{0.000185}[/tex]
= 0.6359
For the percentage of students who favor the fee hike, the 95% confidence interval is (0.5827, 0.6539)
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