A random sample of 400 rails passengers is taken and 55% are in favor of proposed new timetable. Construct a 95% confidence interval that what proportion of all rail passengers is in favor of the proposed timetables.
Answers
Given : A random sample of 400 rails passengers is taken and 55% are in favor of proposed new timetable.
95% confidence interval
To Find : proportion of all rail passengers is in favor of the proposed timetables.
Solution:
n = 400
p = 55% = 0.55
q= 1 - p = 1 - 0.55 = 0.45
Mean = np = 400 * 0.55 = 220
Variance = npq = 220 * 0.45 = 99
SD = √Variance = √99 = 9.95
z for 95 % confidence interval = 1.96
Margin of Error = z * SD for sample / n or z * √(pq/n)
=> Margin of Error = 1.96 * 9.95 / 400 or 1.96 * √(0.55 *0.45/400)
=> Margin of Error = 0.04875
=> Margin of Error = 4.875 %
Hence 55 ± (3 * 4.875) %
rail passengers is in favor of the proposed timetables.
40.375 to 69.625 %
Learn More:
Assume that adults have iq scores that are normally distributed with ...
brainly.in/question/11133397
The mean length of mature whiting is 35 cm and the standard ...
brainly.in/question/13450142
The value of the cumulative standardized normal distribution at z is ...
brainly.in/question/11376268
