AMS 312.01 Practice Final Exam #1 Spring, 2009

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Instructions: This is a close book exam. Anyone who cheats in the exam shall receive a grade of F. Please provide complete solutions for full credit. Good luck!

1. Arctic and Alpine Research investigated the relationship between the mean daily air temperature and the cocoon temperature of woolybear caterpillar’s of the High Arctic.

(a) According to the data, can you conclude, at the significance level of 0.05, that the caterpillar’s body temperature is higher than the outside air temperature?

(b) What assumptions are necessary for the above test?

Temperature (ºC)
Day / Air / Cocoon
1 / 10 / 15
2 / 9 / 14
3 / 2 / 7
4 / 3 / 6
5 / 5 / 10

Solution: By taking the paired differences (Diff) between the cocoon and the air temperatures for each day sampled, this problem reduce to a one-sample t-test on Diff.

Temperature (ºC)
Day / Air / Cocoon / Diff
1 / 10 / 15 / 5
2 / 9 / 14 / 5
3 / 2 / 7 / 5
4 / 3 / 6 / 3
5 / 5 / 10 / 5

(a). Sample statistics: n = 5, s = 0.9. Hypotheses: H0: μ = 0 versus Ha: μ 0.

Test statistic:

Since we reject H0 in favor of Ha at the 0.05 significance level. That is, we conclude, at the significance level of 0.05, that the caterpillar’s body temperature is higher than the outside air temperature.

(b). The assumption is that the population distribution of “Diff” is normal.

2. Let X1, X2, …, Xn be a random sample from a normal population N(μ, σ2). Furthermore, the population variance σ2 is known. For a 2-sided test of H0: μ = μ0 versus Ha: μ ≠ μ0, at the significance level α,

(a). Derive the one-sample z-test using the pivotal quantity method. (* Please include the derivation of the pivotal quantity, the proof of its distribution, and the derivation of the rejection region for full credit.)

(b). Prove that the likelihood ratio test is equivalent to the usual one sample z-test.

Solution:

(a). (1) Since this is inference on one population mean, we start with its point estimator – the sample mean, The distribution of is . The distribution of is not entirely known because μ is unknown. By taking the linear transformation , however, we can see that Z is a function of the sample statistics and the unknown parameter of interest (μ) only. Furthermore the distribution of Z is completely known: . Thus Z is a pivotal quantity for the inference on μ.

(2) The following is the proof of the distribution of Z using the moment generating function method.

Therefore .

(3) Next we derive the one-sample z-test and its rejection region.

For a 2-sided test of H0: μ = μ0 versus Ha: μ ≠ μ0, the test statistic is the pivotal quantity at μ = μ0, that is, . Intuitively, we would reject H0 in favor of Ha if . The problem is how to determine c. By the definition of the significance level, we have

Thus and subsequently we have

That is, at the significance level α, we reject H0 in favor of Ha if .

(b). For a 2-sided test of H0: μ = μ0 versus Ha: μ ≠ μ0, when the population is normal and population variance σ2 is known, we now derive the likelihood ratio test.

(1) Write down your parameter space under

(2) Write down the unrestricted/original parameter space.

(3) Write down the likelihood (of the data)

(4) Write down your log-likelihood.

(5) Find your MLE under and plug it in to your to obtain

(6) Find the MLE(s) under and plug in to your to obtain

(7) Get the likelihood ratio

(8) Derive the decision rule for a given significance level

Recall the z-test we derived before using the Pivotal Quantity method.

Test Statistic :

At , we reject if

3. Let Xi, i = 1, …, n, denote the outcome of a series of n independent trials, where Xi = 1 with probability p, and Xi = 0 with probability (1- p). Let.

(a). Please derive the method of moment estimator of p.

(b). Please derive the maximum likelihood estimator of p.

(c). Please derive the 100(1-α)% large sample confidence interval for p using the pivotal quantity method.

(d). At the significance level α, please derive the large sample test for H0: p = p0 versus Ha: p ≠ p0, using the pivotal quantity method. (* Please include the derivation of the pivotal quantity, the proof of its distribution, and the derivation of the rejection region for full credit.)

Solution:

(a). The population mean is p and the sample mean is . Therefore the moment estimator of p is .

(b). The likelihood function is:

The log likelihood is:

Solving the equation:

, we have

(c). The population distribution is Bernoulli (p), i.e. Xi ~ Bernoulli(p). Therefore the population mean is p and the population variance is p(1-p). When the sample size n is large, by the central limit theorem, we know that the sample mean follows approximately the normal distribution with its mean being the population mean and its variance being the population variance divided by n as follows: .

Same as in 2 (a), we can show that is a pivotal quantity for the inference on p.

We can use this pivotal quantity to construct the large sample confidence interval for p. Alternatively, we can also use the following pivotal quantity to construct the large sample confidence interval as follows.

Therefore the 100(1-α)% large sample confidence interval for p is:

(d). From part (c) above, we have shown that is a pivotal quantity for the inference on p. For a 2-sided test of H0: p = p0 versus Ha: p ≠ p0, the test statistic is the pivotal quantity at p = p0, that is, . Intuitively, we would reject H0 in favor of Ha if . The problem is how to determine c. By the definition of the significance level, we have

Thus and subsequently we have

That is, at the significance level α, we reject H0 in favor of Ha if .

4. John Pauzke, president of Cereals Unlimited Inc., wants to be very certain that the mean weight μ of packages satisfies the package label weight of 16 ounces. The packages are filled by a machine that is set to fill each package to a specified weight. However, the machine has random variability measured by σ2. John would like to have strong evidence that the mean package weight is above 16 ounces. George Williams, quality control manager, advises him to examine a random sample of 25 packages of cereal. From his past experience, George knows that the weight of the cereal packages follows a normal distribution with standard deviation 0.4 ounces. At the significance level α = 0.05:

(a) What is the decision rule (rejection region) in terms of the sample mean?

(b) What is the decision of your test if a sample of 25 packages of cereal yields a mean of 16.3 ounces? What is the p-value of your test?

(c) What is the power of the test when μ= 16.23 ounces? (Please derive the formula for the power of the test first, and then calculate the desired power.)

Solution:

(a) or is a pivotal quantity for this test problem.

If we select Z as the pivotal quantity,

a = P (Reject H0|H0)=P (Z0Za|μ=μ0) =

Therefore we should reject H0 in favor of Ha if Z0 Za where Za =1.645.

From the above, we have . Therefore we should reject H0 in favor of Ha if C* where C* =16.13.

(b) Since 16.3>16.13, we should reject H0 in favor of Ha. P-Value is the tail area.

. Therefore P-Value = 0.0001.

====1-0.1093=0.8907

5. A new method of making concrete blocks has been proposed. To test whether or not the new method increases the compressive strength, five sample blocks are made by each method. The compressive strengths in 10 pounds per square inch are listed here:

New Method / 15 14 13 15 16
Old Method / 13 15 13 12 14

Suppose that both populations are normally distributed and furthermore, (although they are unknown).

(a). Please construct a 95 % percent confidence interval for the mean difference between the compressive strengths by two methods.

(b). At the significance level .05, can you conclude that the new method increases the compressive strength?

Solution: Inference on two population means. Two small and independent samples.

New method:

Old method:

(a) 95% C. I. for difference is

where

Therefore 95% C.I. is [-0.46, 2.86].

(b) Using t-test with hypotheses v.s. ,

Thus, we cannot reject H0. There isn't enough evidence to conclude that the new method increases the strength.