WILL GIVE BRAINLIEST AND 100 POINTS
Please help, I need to find this question out but I can not understand it.
A recent survey of 128 high school students indicated the following information about release times from
school.
-35 percent wanted to start school and hour earlier and get out earlier
-20 percent liked things how there are currently
-39 percent wanted to start school a half an hour later and get out later in the day.
-6 percent had no opinion
The margin of error for this survey was calculated to be plus or minus 4 percent.
The sample size n=128
Find the interval estimates for those that want to get out earlier and those that want to get out later.
Explain how it would be possible for the group that wants to get out of school earlier to win after all
the votes are counted.

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Respuesta :

tomdodd2003 tomdodd2003 · Answer 1 · 2020-05-13T20:27:42+02:00

Answer:

a) The interval for those who want to go out earlier is between 43.008 and 46.592

b) The interval for those who want to go out later is between 47.9232 and 51.9168

Step-by-step explanation:

Given that:

Sample size (n) =128,

Margin of error (e) = ±4% =

a) The probability of those who wanted to get out earlier (p) = 35% = 0.35

The mean of the distribution (μ) = np = 128 * 0.35 = 44.8

The margin of error = ± 4% of 448 = 0.04 × 44.8 = ± 1.792

The interval = μ ± e = 44.8 ± 1.792 = (43.008, 46.592)

b) The probability of those who wanted to start school get out later (p) = 39% = 0.39

The mean of the distribution (μ) = np = 128 * 0.39 = 49.92

The margin of error = ± 4% of 448 = 0.04 × 49.92 = ± 1.9968

The interval = μ ± e = 44.8 ± 1.792 = (47.9232, 51.9168)

The way for those who want to go out earlier to win if the vote is counted is if those who do not have any opinion vote that they want to go earlier