Pset 1

assuming you already find a,b.. i

Assuming independence:

$$P(X=x_i,y=y_i)=P(X)P(Y)$$

or

We know the cond probability

x = 1 given y=10 is 0.5

does not equal

P(x=1)=G

Not independent

Compute the mean is easy after doing part 1

$$P(x=1)=G$$ $$P(x=1)=0.2$$ $$P(x=3)=I$$ $${\mu }_x+E(x)=\sum x_{i}P(x=x_i)$$ $$=1\cdot P(x=1)+2\cdot p(x=2)$$ $$=?+3\cdot P(x=3)$$ $$E(x^2)=\dots$$

The variance of x

$${\sigma }_x^2=E(x^2)-{\mu }_x^2=?$$ $$E(x|y=6)$$ $$=\sum x_i:P(x=x_i|y=0)$$ $$=1\cdot P(x=1|y=6)+2\cdot P(x=2|y=6)+3\cdot P(x=3|y=6)$$ $$=?$$ $$E(x|y=6)$$ $$E(x|y=8)$$ $$E(x|y=10)$$

if these three are the same, then x is mean independent of y

part d: if it is mean independence, then the covariance is 0

$$Cov(x,y)=\sum x_i{y}_j$$

Problem 2:

Define

$$x_i=\{\begin{array}{ll}1& \text{ if the i’th mussel is good}\\ 0& \text{ if not}\end{array}$$ $$x_i\approx Bernoulli(P)$$

where

$$P=P(X=1)=0.9$$ $$\sum X_i=\text{ The number of good nussels in a sample }$$ $$P\left(\sum X_i<6\right)$$ $$=P\left(\frac{\sum x_i-60\cdot P}{\sqrt{60\cdot p\cdot (1-p)}}<\frac{6-60\cdot p}{\sqrt{60\cdot p\cdot (1-p)}}\right)$$ $$\approx P\left(Z<\frac{6-60\cdot p}{\sqrt{60\cdot P(1-P)}}\right)$$

see photo on phone

Problem #3

N children in small town.

Define buronelli variable x for cavities or no cavities

$$i=1,\dots n$$ $$x_i{\approx }^{iid}Bernoulli(P)$$

when $P=P(X=1)$ Then, $n_1=\sum (x_i)$ This is the total number of children with cavities in a sample of size n.

A: is the sample average unbiased?

$$\frac{n_1}{n}=\frac{1}{n}\sum x_i$$

Apply large number thing

$$E\left(\frac{n_1}{n}\right)=E\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)=P$$

Not biased

ask kyle for picture

B: assume total population 200

Sample size 80

Population size is 150/200

true p

$$p=\frac{150}{200}$$

Find probability that more then 55 children have cavities in samp size of 80

$$P\left(\sum _{i=1}^{80}{x}_i>55\right)$$ $$=P\left(\frac{{\sum }^{80}{x}_i-80\cdot p}{\sqrt{80\cdot p\cdot (1-p)}}>\frac{55-80\cdot P}{\sqrt{80\cdot P()}}\right)$$

See kyle picture

C: find 90% confidence interval

B and c are not related. We do not know the true proportion, only sample estimate

$$\left(\hat{P}-Z_{\frac{\alpha }{2}}\cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right)$$

Where n is 80

$$\hat{p}+z_{\frac{\alpha }{2}}\cdot \sqrt{\left(\frac{\hat{p}(1-\hat{p})}{n}\right)}$$

D :

$$p=0.75vs{H}_1=P\ne 0.75$$

Under $H_0$

$$z=\frac{\hat{p}-p}{\sqrt{{\hat{p}}_1-\hat{p}}}$$

follows approximately standard normal

$$N(0,1)$$ $$z_{act}={\hat{p}}_{\text{actual}}-p..$$

Problem 3 Revisited:

N is total population of children n is the sample size (small n)

$n_1$ is the number of children that had cavities in the sample size of n students.

Population proportion is $\propto$ to population mean?

Define

$$x=\{\begin{array}{ll}1& \text{ if child has cavities}\\ 0& \text{otherwise }\end{array}$$

x is a Bernoulli variable with proportion p

p is the proportion of children that have cavities, or the probability that a child has a cavity

Now:

$$\sum _{i=1}^n{x}_i=n_1$$

The mean of the burnoulli random var is equal to P Variance of the bernoulli random variable P(1-P)

By wigglos large number:

$$\frac{n_1}{n}=\frac{1}{n}\sum _{i=1}^n{x}_i\to p\text{ as }n\to \infty$$

Unbiased! as long as $\frac{n_1}{n}$ is a sample average by wigglos large number, then it is unbiased

B: The true proportion is given by $\frac{150}{200}=0.75$ Find the probability that the sample has more then 55 children with cavities

$$P\left(\sum x_i>55\right)$$

Standardize $x_i$ by dividing mean by standard deviation

mean is given by

$$E\left(\sum ^{80}{x}_i\right)=80\ast p=80\ast 0.75$$

Variance

$$Var\left(\sum ^{80}{x}_i\right)=n-p(1-p)$$ $$=80\ast 0.75(0.25)$$

=b?

central limit theorem, the lhs of picture follows it

Part C: You dont need the 200

Sample average is $\hat{p}=\frac{n_1}{n}$

$$\hat{p}-z_{\frac{\alpha }{2}}\cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}},\hat{p}+z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{2}}$$