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2 answers:
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Answer:
a)
b)
Step-by-step explanation:
1) Data given and notation
n=900 represent the random sample taken
X=372 represent the students were pursuing liberal arts degrees
estimated proportion of students were pursuing liberal arts degrees
represent the significance level
z would represent the statistic (variable of interest)
represent the p value (variable of interest)
p= population proportion of students were pursuing liberal arts degrees
2) Solution to the problem
The confidence interval would be given by this formula
For the 99% confidence interval the value of and
, with that value we can find the quantile required for the interval in the normal standard distribution.
The margin of error is given by:
If we replace we have:
And replacing into the confidence interval formula we got:
And the 99% confidence interval would be given (0.371;0.455).
We are confident (99%) that about 37.1% to 45.5% of students were pursuing liberal arts degrees.