Problem 1.

Consider a disease with two different phases. In a population there are $75 \%$ of people who are disease free, $25 \%$ of people have the disease. Among those with the disease, $80 \%$ have a latent form of the disease and $20 \%$ have the manifest form. There is a diagnostic test for the disease. The test will be positive in $90 \%$ of people who have the manifest form, and in $70 \%$ of people who have the latent form. The test will also give a false positive result in $20 \%$ of people who do not have the disease.

(a) For a randomly selected person who has the disease (in either the latent or manifest form) what is the probability that the person has a positive test result?

(b)If a person chosen at random from the entire population has a positive test result, what is the probability that this person has the disease (in either the latent or manifest form)?

Proof .

Define the events $D=$ has disease, $P T=$ positive test, $L=$ person has latent form of disease, $M=$ person has manifest form. Then, since we know that the person has the disease, we can use the law of total probability and split people into groups with latent or manifest forms:
$$P(P T \mid D)=P(P T \mid L) P(L \mid D)+P(P T \mid M) P(M \mid D)=0.7 \cdot 0.8+0.9 \cdot 0.2=0.74$$

Now we need another event: $H=$ person is healthy. Then, considering the whole population
$$P(H)=0.75, \quad P(L)=0.25 \cdot 0.8=0.2, \quad P(M)=0.25 \cdot 0.2=0.05$$
Use Bayes’ rule and the result obtained above to find
$$\begin{gathered} P(D \mid P T)=\frac{P(P T \mid D) P(D)}{P(P T)}=\frac{P(P T \mid D) P(D)}{P(P T \mid H) P(H)+P(P T \mid D) P(D)} \\ =\frac{0.74 \cdot 0.25}{0.2 \cdot 0.75+0.74 \cdot 0.25}=0.552 \end{gathered}$$

Problem 2.

You wish to test whether your cat prefers one paw over the other. You dangle a ribbon in front of your cat ten times, and he bats it with his right paw 8 of those times and with his left paw 2 times. Use an exact Binomial test at significance level $\alpha=0.05$.

Proof .

(a) Carefully define the population parameter involved in this test in words.
Let $p$ denote the proportion of times (ever, not just on that day) that your cat bats with his right paw.
(b) Define the null hypothesis and alternative for the test using the population parameter you defined above.
$$H_{0}: p=\frac{1}{2}, \quad H_{a}: p \neq \frac{1}{2}$$
(c) State the value of the test statistic and the distribution of the test statistic under the null hypothesis.

The test statistic is $X=$ number of times (out of 10 trials) that the cat batted with the right paw. Then $X \sim \operatorname{Binomial}\left(n=10, p=\frac{1}{2}\right)$. Here $X=8$
(d) Compute the $p$-value of this test and formulate a conclusion sentence.
The $p$-value is the probability to observe values of the test statistic as unusual or more unusual than the one observed if the null hypothesis were true. We observed $X=8 .$ More unusual would be $X=9$ or $X=10$, but also $X=0,1,2 .$ The Binomial distribution with $p=\frac{1}{2}$ is symmetric, so
\begin{aligned} &p=P(X \in\{0,1,2,8,9,10\})=2 P(X \in\{8,9,10\}) \\ &=2 \cdot\left[\left(\begin{array}{c} 10 \\ 8 \end{array}\right)\left(\frac{1}{2}\right)^{10}+\left(\begin{array}{c} 10 \\ 9 \end{array}\right)\left(\frac{1}{2}\right)^{10}+\left(\begin{array}{c} 10 \\ 10 \end{array}\right)\left(\frac{1}{2}\right)^{10}=0.109\right] \end{aligned}
Since this $p$-value is large (larger than $\alpha=0.05)$, we fail to reject the null hypothesis. There is not enough evidence in this experiment to conclude that the cat prefers one paw over another.

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