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2 years ago

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Suppose that we were going to conduct a study to determine if there is an increased risk of getting lung cancer based on whether

a person smokes or not. In the group of smokers(1) , we find that 10 out of 100 get lung cancer. In the non-smokers group(2), we find that 5 out of 200 get lung cancer. What is the sample relative risk?

1 answer:

oee [108]2 years ago

4 0

Answer:

Sample relative risk is 4.

Step-by-step explanation:

We are given the following in the question:

Group 1 of smokers: 10 out of 100 get lung cancer.

$\text{P(Lung cancer for smoker)} = \dfrac{10}{100} = \dfrac{1}{10}$

Group 2 of non-smokers: 5 out of 200 get lung cancer

$\text{P(Lung cancer for non-smoker)} = \dfrac{5}{200} = \dfrac{1}{40}$

Sample relative risk:

It is the ratio of probability of an outcome in an exposed group to the probability of an outcome in an unexposed group.

Sample relative risk =

$\dfrac{\text{P(lung cancer for smokers)}}{\text{P(lung cancer for non smokers)}}\\\\=\dfrac{\frac{1}{10}}{\frac{1}{40}} = \dfrac{40}{10} = 4$

Thus, sample relative risk is 4.

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Solve the following equations: (a) x^11=13 mod 35 (b) x^5=3 mod 64

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a.

$x^{11}=13\pmod{35}\implies\begin{cases}x^{11}\equiv13\equiv3\pmod5\\x^{11}\equiv13\equiv6\pmod7\end{cases}$

By Fermat's little theorem, we have

$x^{11}\equiv (x^5)^2x\equiv x^3\equiv3\pmod5$

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5 and 7 are both prime, so $\varphi(5)=4$ and $\varphi(7)=6$ . By Euler's theorem, we get

$x^4\equiv1\pmod5\implies x\equiv3^{-1}\equiv2\pmod5$

$x^6\equiv1\pmod7\impleis x\equiv6^{-1}\equiv6\pmod7$

Now we can use the Chinese remainder theorem to solve for $x$ . Start with

$x=2\cdot7+5\cdot6$

Taken mod 5, the second term vanishes and $14\equiv4\pmod5$ . Multiply by the inverse of 4 mod 5 (4), then by 2.

$x=2\cdot7\cdot4\cdot2+5\cdot6$

Taken mod 7, the first term vanishes and $30\equiv2\pmod7$ . Multiply by the inverse of 2 mod 7 (4), then by 6.

$x=2\cdot7\cdot4\cdot2+5\cdot6\cdot4\cdot6$

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b.

$x^5\equiv3\pmod{64}$

We have $\varphi(64)=32$ , so by Euler's theorem,

$x^{32}\equiv1\pmod{64}$

Now, raising both sides of the original congruence to the power of 6 gives

$x^{30}\equiv3^6\equiv729\equiv25\pmod{64}$

Then multiplying both sides by $x^2$ gives

$x^{32}\equiv25x^2\equiv1\pmod{64}$

so that $x^2$ is the inverse of 25 mod 64. To find this inverse, solve for $y$ in $25y\equiv1\pmod{64}$ . Using the Euclidean algorithm, we have

64 = 2*25 + 14

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14 = 1*11 + 3

11 = 3*3 + 2

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=> 1 = 9*64 - 23*25

so that $(-23)\cdot25\equiv1\pmod{64}\implies y=25^{-1}\equiv-23\equiv41\pmod{64}$ .

So we know

$25x^2\equiv1\pmod{64}\implies x^2\equiv41\pmod{64}$

Squaring both sides of this gives

$x^4\equiv1681\equiv17\pmod{64}$

and multiplying both sides by $x$ tells us

$x^5\equiv17x\equiv3\pmod{64}$

Use the Euclidean algorithm to solve for $x$ .

64 = 3*17 + 13

17 = 1*13 + 4

13 = 3*4 + 1

=> 1 = 4*64 - 15*17

so that $(-15)\cdot17\equiv1\pmod{64}\implies17^{-1}\equiv-15\equiv49\pmod{64}$ , and so $x\equiv147\pmod{64}\implies\boxed{x\equiv19\pmod{64}}$

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