1. A college admissions officer for the school’s online undergraduate program wa
ID: 3130409 • Letter: 1
Question
1. A college admissions officer for the school’s online undergraduate program wants to estimate the mean age of its graduating students. From a previous study the standard deviation was approximately two (2) years.
A. The administrator took a random sample of 40 from which the mean was 24 years and the standard deviation was 1.7 years. From this sample, what is the 95 confidence interval for the population mean of interest?
B. What is a proper statistical interpretation of the confidence interval you calculated in Part B?
C. Would a 95% confidence interval using a larger sample size, assuming the same sample mean and standard deviation, be wider or narrower than the interval using the administrator’s sample size? Explain why it would be wider or narrower.
D. Typically, undergraduate resident students graduate by the time they are 23 years of age. Does the interval you calculate in Part B reflect that the average graduating student age of the online student is older than that of the graduating resident student? Explain.
Explanation / Answer
a)
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 24
z(alpha/2) = critical z for the confidence interval = 1.959963985
s = sample standard deviation = 1.7
n = sample size = 40
Thus,
Margin of Error E = 0.526825777
Lower bound = 23.47317422
Upper bound = 24.52682578
Thus, the confidence interval is
( 23.47317422 , 24.52682578 ) [ANSWER]
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b)
We are 95% confident that the true mean age of online graduating students is between 23.473 and 24.527 years. [ANSWER]
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c)
It will be narrower, as the margin of error is inversely proportional to the square root of the sample size.
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d)
YES, because the whole interval is greater than 23.
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