question archive You have been given the task of finding out what proportion of students that enroll in a local university actually complete their degree
You have been given the task of finding out what proportion of students that enroll in a local university actually complete their degree
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You have been given the task of finding out what proportion of students that enroll in a local university actually complete their degree. You have access to first year enrolment records and you decide to randomly sample 118 of those records. You find that 87 of those sampled went on to complete their degree.
a)Calculate the proportion of sampled students that complete their degree. Give your answer as a decimal to 3 decimal places.
Sample proportion =
You decide to construct a 95% confidence interval for the proportion of all enrolling students at the university that complete their degree. You may find this standard normal table useful for the following questions. If you use your answer to part a) in the following calculations, use the rounded version.
b)Calculate the lower bound for the confidence interval. Give your answer as a decimal to 3 decimal places.
Lower bound for confidence interval =
c)Calculate the upper bound for the confidence interval. Give your answer as a decimal to 3 decimal places.
Upper bound for confidence interval =
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a)Sample proportion is=0.737
The 95% confidence interval is (0.658, 0.816)
b)Lower bound for confidence interval =0.658
c)Upper bound for confidence interval = 0.816
Step-by-step explanation
You have been given the task of finding out what proportion of students that enroll in a local university actually complete their degree. You have access to first year enrollment records and you decide to randomly sample 118 of those records. You find that 87 of those sampled went on to complete their degree.
a)Sample proportion is;
x=87
n=118
p?=x/n
p?=87/118=0.737
Therefore the sample proportion is=0.737
A 95% confidence interval for the proportion of all enrolling students at the university that complete their degree is constructed as follows;
At 95% confidence level, α=1-0.95=0.05
zc=1.96
?C.I=p^?±zc?⋅np^?(1−p^?)???
?=0.737±1.96⋅1180.737(1−0.737)???
?=0.737±1.96⋅1180.737(0.263)???
?=0.737±0.079?
The 95% confidence interval is (0.658, 0.816)
b)Lower bound for confidence interval =0.737-0.079=0.658
c)Upper bound for confidence interval =0.737+0.079=0.816
