5. (14 pts) (sec 6.3 #10) sampling distribution: use the population {4, 5, 9} that was used in the textbook and classroom examples. also, as in the text and in class, assume samples of size n = 2 are randomly selected with replacement. hint: the proportion of odd numbers in the population is 2/3. a) for the population, find the proportion of odd numbers (in each sample). you need to fill in the following table first. b) find the mean of the sampling distribution of the sample proportion of odd numbers. c) based on the results, is the sample proportion and unbiased estimator of the population proportion? why or why not?

Question

Accepted Answer

# Explanation:
## Step1: Calculate proportion for each sample
- For sample $4,9$: There is $1$ odd number ($9$), so proportion $=\frac{1}{2} = 0.5$
- For sample $5,4$: There is $1$ odd number ($5$), so proportion $=\frac{1}{2}=0.5$
- For sample $5,5$: There are $2$ odd numbers, so proportion $=\frac{2}{2} = 1$
- For sample $5,9$: There are $2$ odd numbers, so proportion $=\frac{2}{2}=1$
- For sample $9,4$: There is $1$ odd number ($9$), so proportion $=\frac{1}{2}=0.5$
- For sample $9,5$: There are $2$ odd numbers, so proportion $=\frac{2}{2}=1$
- For sample $9,9$: There are $2$ odd numbers, so proportion $=\frac{2}{2}=1$

## Step2: Calculate probability for each sample
Since samples are selected with replacement, total number of samples $n = 3\times3=9$. Probability of each sample $=\frac{1}{9}$

## Step3: Calculate mean of sampling distribution ($\mu_{\hat{p}}$)
$$
\begin{align*}
\mu_{\hat{p}}&=\sum(\hat{p}\times P(\hat{p}))\
&=(0\times\frac{1}{9})+(0.5\times\frac{4}{9})+(1\times\frac{4}{9})\
&=0+\frac{2}{9}+\frac{4}{9}\
&=\frac{2 + 4}{9}\
&=\frac{6}{9}=\frac{2}{3}
\end{align*}
$$

## Step4: Check if sample proportion is unbiased estimator
Population proportion $p=\frac{2}{3}$. Since $\mu_{\hat{p}}=\frac{2}{3}=p$, the sample proportion is an unbiased estimator.

# Answer:
a)
|Sample|Proportion|Probability|
|---|---|---|
|4,4|0|$\frac{1}{9}$|
|4,5|0.5|$\frac{1}{9}$|
|4,9|0.5|$\frac{1}{9}$|
|5,4|0.5|$\frac{1}{9}$|
|5,5|1|$\frac{1}{9}$|
|5,9|1|$\frac{1}{9}$|
|9,4|0.5|$\frac{1}{9}$|
|9,5|1|$\frac{1}{9}$|
|9,9|1|$\frac{1}{9}$|

b) $\frac{2}{3}$

c) Yes, because the mean of the sampling distribution of the sample proportion ($\mu_{\hat{p}}=\frac{2}{3}$) is equal to the population proportion ($p = \frac{2}{3}$).

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate proportion for each sample

Step2: Calculate probability for each sample

Step3: Calculate mean of sampling distribution (\(\mu_{\hat{p}}\))

Step4: Check if sample proportion is unbiased estimator

Answer:

Sample	Proportion	Probability
4,5	0.5	\(\frac{1}{9}\)
4,9	0.5	\(\frac{1}{9}\)
5,4	0.5	\(\frac{1}{9}\)
5,5	1	\(\frac{1}{9}\)
5,9	1	\(\frac{1}{9}\)
9,4	0.5	\(\frac{1}{9}\)
9,5	1	\(\frac{1}{9}\)
9,9	1	\(\frac{1}{9}\)