STAT415 Assessment #5 Exam
You must show all of your work in order to receive full and/or partial credit.
1. 6pts—Let X and s2 denote the mean and variance for a random sample of size 10 taken from a normal population with mean μ. Find a one-sided confidence interval for μ of the form (c,∞). In other words, find c such that P(c < μ) = .95. 2. For fixed θ > 0, let X1,…,Xn be a random sample from a population with pdf f(x) = θ(1−x)θ−1, 0 ≤ x ≤ 1.
and expectation E(X) = 1 . 1+θ
(a) 4pts—Find the method of moments estimator for θ.
(b) 4pts—Find the maximum likelihood estimator (MLE) of θ.
3. The following data summarizes the responses for 200 randomly selected subscribers to a certain magazine. They were asked their age and whether they have a Facebook account.
Age group Yes No
18-49 50+
96 24 49 41
(a) 4pts—For these 200 individuals, what is the difference between the proportion “Yes” for the 18-49 age group and the proportion “Yes” for the 50+ group? Add (and subtract) a 95% margin of error to this estimate to form a confidence interval.
(b) 4pts—Provide an interpretation of this interval. Specifically, explain briefly why we need an interval estimate and not simply the sample proportion by itself.
4. From each in a sample of 737 women aged 25-50 years old, daily calcium (y) and iron (x) intakes were measured (both in mg).
(a) 4pts—Explain briefly why one cannot use such data to estimate the difference between the mean calcium and mean iron intake with the formula
√
( y − x ) ± 2 s 2y + s 2x 737 737
(b) 4pts—The regression model for these variables was estimated with the following equa- tion:
yˆ = 624.05 + 26.252(x − x)
In terms of calcium and iron, how is the value 26.252 interpreted?
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