29. The candy machine Suppose a large candy machine has 45% orange candies. Imagine taking an SRS of 25 candies from the machine and observing the sample proportion $\hat{p}$ of orange candies.
(a) What is the mean of the sampling distribution of $\hat{p}$? Why?
(b) Find the standard deviation of the sampling distribution of $\hat{p}$. Check to see if the $10\%$condition is met.
(c) Is the sampling distribution of $\hat{p}$approximately Normal? Check to see if the Normal condition is met.
(d) If the sample size were 50 rather than 25, how would this change the sampling distribution of $\hat{p}$?

To determine the mean of the sampling distribution of $\hat{p}$.

Let the sample distribution to be $p$and the $SRS$sample size to be $n$.
$p=45\%$
$=0.45$
Then, $n=25$

The mean of a sample proportion's sampling distribution $\hat{p}$and the population proportion $p$are equivalent.

${\mu }_{\hat{p}}=p$.

Let, substitute the value $0.45$for $p$as:

The mean is calculated using the sampling proportion as an unbiased estimator of the population percentage.

As a result, the mean of the sampling distribution $\hat{p}$is $0.45$.

To find the standard deviation of the sampling distribution of $\hat{p}$. Then to check to see if the $10\%$ condition is met.

Let, the sample distribution to be $p$and the sample size for $SRS$to be $n$.
$p=45\%$

$=0.45$

Then, $n=25$

The standard deviation of the sampling distribution of $\hat{p}$is ${\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$.

Let, substitute the value $0.45$for $p$.

Also substitute the value, $25$for $n$ .
${\sigma }_{\hat{p}}=\sqrt{\frac{0.45(1-0.45)}{25}}$
$=\sqrt{\frac{0.45\times 0.55}{25}}$
$=\sqrt{0.0099}$

$\approx 0.0994987$

The candy machine is enormous, it contains more than $250$candies, so satisfying the $10\%$ criterion.

As a result, the standard deviation is $0.099487.$

To find the sampling distribution of $\hat{p}$ is approximately normal or not.

The sampling distribution is roughly Normal when the product of sample size and sampling proportion, that is, $np$and $n(1-p)$, are both less than or equal to $10$.
Let, the sample distribution to be $p$and sample size for $SRS$to be $n$.
$p=45\%$
$=0.45$
Then,$n=25$
If the product of the sample size and the sampling proportion, $np$and $n(1-p)$, are both less than $10$, the sampling distribution is nearly Normal.

Let, substitute the value$0.45$for $p$and the value $25$for $n$in the term $np$.
$25\times (0.45)$

$=11.25$

Then, substitute the value $0.45$for $p$and the value $25$for $\mathrm{n}$in the term $n(1-p)$.
$25(1-0.45)=25\times 0.55$
$=13.75$
Because both $np$and $n(1-p)$are at least $10$ the Normal distribution requirement is met.
As a result, the sampling distribution of $\hat{p}$ is approximately Normal.

The sample size were $50$rather than $25$, and change the sampling distribution of $\hat{p}$.

Let, the sample distribution to be $p$and the sample size for $SRS$to be $n$.
$p=45\%$

$=0.45$

$n=50$

The standard deviation of the sampling distribution of $\hat{p}$is calculated by:${\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$.

Then, substitute the value $0.45$for $p$and the value $50$for $n$as:

${\sigma }_{\hat{p}}=\sqrt{\frac{0.45(1-0.45)}{50}}$

$=\sqrt{\frac{0.45\times 0.55}{50}}$

$=\sqrt{0.00495}$

$\approx 0.0703562$

As a result, the standard deviation is $0.0703562$.