ISYE3039 HW2 Solutions

ISyE 3039 – HW2 Solutions

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Problem 1:


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Problem 2:

ISy

of

Problem

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m 3:

W2 Solutions

ISy

of

Use α

yE 3039 – HW

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α=0.05. Wha

W2 Solutions

t are the pra

actical implica

ations if the nnull hypothessis is rejectedd?


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Problem 4: This problem is based on 4.28 but H1 is different from the textbook

Suppose we wish to test the hypotheses

: 15

: 15

where we know that 9.0. If the true mean is really 20, what sample size must

be used to ensure that the probability of type II error is no greater than 0.10?

Assume that 0.05.

The sample mean has the distribution: ~ 20, , but test statistic is /

.

Then type II error is

Pr15

9/ . Pr20

9/1.645

59/

Φ 1.645 1.667√ 0.1

From Normal CDF table, we have 1.645 1.667√ 1.29

or 3.10

Therefore, the sample size should be 4 or larger.


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Problem 5

Another way is to compute the test statistic based on the sum, i.e. on ∑ ~POI , but in this way, the Poisson distribution is approximated by , , not , / . The test

statistic is ∑ / . Reject : if | | /

This two methods are equivalent. For example, if we use data of this problem, then we have

11, 100, 0.15

And we can compute that √

1.0328 which is the same as method 1.


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Problem 6: Extra credits

If the sample sizes of two populations are equal (denoted as n), then we can develop confidence interval based on t-statistics for , which is supposed to be normally distributed with mean and variance .

So we can pair the individual measurements from the two groups and get sampled values of , and compute their mean x and standard deviation s , so the confidence interval is

/ , √ / , √

Since the individual measurements within a group are independent to each other. We do not care about how to pair.

If the sample sizes of two populations are not equal, then we cannot apply such a simple method but need Equations 4.57 for confidence interval:

where degrees of freedom v is given by Equation 4.55

ISYE3039 HW2 Solutions

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