Math Biostatistics Boot Camp 1

Mathematical Biostatistics Boot Camp 1

Week 3 Homework

1. (1 point) A web site (www.medicine.ox.ac.uk/bandolier/band64/b64-7.html) for home pregnancy tests cites the following: “When the subjects using the test were women who collected and tested their own samples, the overall sensitivity was 75%. Specificity was also low, in the range 52% to 75%.” Suppose a subject has a negative test. Assume the lower bound for the specificity. What number is closest to the multiplier of the pre-test odds of pregnancy to obtain the post-test odds of pregnancy given a negative test result? b. 0.5

a. 1.5

c. 2

d. 1 1.

b.

Solution: 1 − sensitivity 1 − 0.75 0.25 DLR− = = = = 0.4808 ≈ 0.5 specificity 0.52 0.52

2. (1 point) A web site (www.medicine.ox.ac.uk/bandolier/band64/b64-7.html) for home pregnancy tests cites the following: “When the subjects using the test were women who collected and tested their own samples, the overall sensitivity was 75%. Specificity was also low, in the range 52% to 75%.” Assume the lower value for specificity. Suppose a subject has a negative test and that 30% of women taking pregnancy tests are actually pregnant. What number is closest to the probability of pregnancy given a negative test? a. 30%

b. 60%

c. 90%

d. 20%

e. 80%

f. 50%

g. 40%

h. 70%

i. 10% 2.

d.

Solution: P (−|preg)P (preg) P (−|preg)P (preg) + P (−|pregc )P (pregc ) (1 − P (+|preg))P (preg) = (1 − P (+|preg))P (preg) + P (−|pregc )(1 − P (preg)) (1 − 0.75)(0.30) 0.075 = = = 0.1708 ≈ 20% (1 − 0.75)(0.30) + (0.52)(0.7) 0.439

P (preg|−) =

3. (1 point) Suppose that hospital infection counts are models as Poisson with mean µ. Recall the Poisson mass µx e−µ for x = 0, 1, . . . Three independent hospitals are observed for one year and their function with mean µ is x! infection counts were 5, 4, and 6, respectively. What is the ML estimate for µ? a. 0

b. 1

c. 4

d. 4.5

e. 5

f. 5.5

g. 6 3.

Solution: 3 Y µxk e−µ i=1

xk !

=

e.

µx1 e−µ µx2 e−µ µx3 e−µ µx1 +x2 +x3 e−3µ µ5+4+6 e−3µ µ15 e−3µ = = = x1 !x2 !x3 ! x1 !x2 !x3 ! x1 !x2 !x3 ! x1 !x2 !x3 !

We will take the derivative of this function and set it to zero. It will be easier first to take logarithm of the function and then find the maximum: f˚ = log f = 15 log(µ) − 3µ − log(x1 !x2 !x3 !) d 15 Now, we find the maximum: (15 log(µ) − 3µ − log(x1 !x2 !x3 !)) = −3 = 0 and solve for µ to get µ = 5. dµ µ

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4. (1 point) Let X1 , . . . , Xn be iid exponential(β). That is, having density

1 −x/β e for x > 0. What is the ML β

estimate for β? a.

n X xi i=1

n

b.

n X

xi

i=1

c.

n X log(xi ) i=1

!−1 d.

n

n X i=1

!−1 xi

e.

n X log(xi ) i=1

n

n X xi

f.

i=1

4.

!−1

n a.

Solution: n Pn Y 1 1 1 −xi /β 1 e = n e−(x1 +···+xn )/β = n e− β i=1 xi β β β i=1 We will take the derivative of this function and set it to zero. It will be easier first to take logarithm of the n 1X function and then find the maximum: f˚ = log f = − xi − n log β β i=1 ! n n d 1X 1 X n − xi − n log β = 2 xi − = 0 and solve for β to get Now, we find the maximum: dβ β i=1 β i=1 β β=

n X xi i=1

n

5. (1 point) Let X be a geometric random variable. That is X counts the number of coin flips until one obtains the first head. The mass function is P (X = x) = p(1 − p)x−1 for x = 1, 2, . . . What is the maximum likelihood estimate for p if one observes a geometric random variable? a. 1/(x − 1)

b. 1/x

c. 1/2

d. 1/(x + 1) 5.

b.

Solution: We want to find the derivative of this function and set it to zero. It will be easier to take the logarithm of the function first: log f = log(p) + (x − 1) log(1 − p) and now we take the derivative: d 1 x−1 (log(p) + (x − 1) log(1 − p)) = − =0 dp p 1−p 1 and solve for p to get p = . x

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6. (1 point) Let X be a Poisson count with mean µ. Recall the Poisson mass function with mean µ is x = 0, 1, . . . What is the maximum likelihood estimate for µ? a. µ

b. 1/x

c. 1/2

d. x2

µx e−µ for x!

e. x 6.

e.

Solution: First, we take the logarithm of the function to get: log f = x log(µ) − µ − log(x!) now we take the derivative and set it to zero to get maximum: x d (x log(µ) − µ − log(x!)) = − 1 = 0 dµ µ and solve for µ to get µ = x.

Question:

1

2

3

4

5

6

Total

Points:

1

1

1

1

1

1

6

Score:

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