Q.30 - The Practice of Statistics for AP

Question

The candy machine Suppose a large candy machine has $15\%$ 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 $75$ rather than $25$, how would this change the sampling distribution of $\hat{p}$ 

Step-by-Step Solution

Answer

a). The required mean is  $0.15$.

b). The required standard deviation is $0.0714143$.

c). $\hat{p}$ sample distribution is not close to Normal.

d). The standard deviation is $0.57445$.

Part (a) Step 1: Given Information

Given in the question that,  a large candy machine has$15$ percent of orange candies

$SRS$$=25$candy count.

Part (a) Step 2: Explanation

Assume that the sample size for SRS is $n$ and that the sample distribution is $p$.

So, $n=25$ and

$p=15\%$

$=0.15$

The mean of a sample proportion's sampling distribution $\hat{p}$ and the population proportion $p$ are the same, i.e., ${\mu }_{\hat{p}}=p$.

In the given formula, replace $p$ with $0.15$.

${\mu }_{\hat{p}}=0.15$

The mean is ${\mu }_{\hat{p}}=0.15$ since the sampling proportion is an unbiased estimate for the population proportion.

Part (b) Step 1: Given Information

$15$ percent  of orange candies in the machine.

SRS $=25$ candy count.

Part (b) Step 2: Explanation

Assume the sample distribution is $p$ and the sample size is $n$.

$p=15\%$

$=0.15$

And, $n=25$

Determine the standard deviation of the sampling distribution:

$\hat{p}$ is ${\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$

Substitute $0.15$ for $p$ and $25$ for $\mathrm{n}$ :

${\sigma }_{\hat{p}}=\sqrt{\frac{0.15(1-0.15)}{25}}$

$=\sqrt{\frac{0.15\times 0.85}{25}}$

$=\sqrt{0.0051}$

$\approx 0.0714143$

Part (c) Step 1: Given Information

$15$ percent  of orange candies in the machine.

SRS $25$ candy count.

Part (c) Step 2: Explanation

Assume that the sample distribution be $p$ and sample size for SRS be $n$.

$p=15\%$

And, $n=25$

The product of sample size and the sampling proportion that is, $np$ and $n(1-p)$ both are less than at least $10$ then the distribution of the samples is roughly Normal.

In the expression $np$, substitute $0.15$ for $p$ and $25$ for $n$.

$25\times (0.15)=3.75$

Part (c) Step 3: Explanation

In the expression $n(1-p)$, substitute $0.15$ for $p$ and $25$ for $n$.

$25(1-0.15)=25\times 0.85$

                     $=21.75$

Both $np$ and $n(1-p)$ are fewer than $10$, indicating that the Normal distribution requirement has not been satisfied.

As a result, $\hat{p}$'s sampling distribution is not close to Normal.

Part (d) Step 1: Given Information

$15$ percent  of orange candies in the machine.

SRS $25$ candy count.

Part (d) Step 2: Explanation

Assume the sample distribution is $p$ and the SRS sample size is$n$.

$p=15$

And, 

$n=75$

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

Simplify the preceding equation by substituting $0.15$ for $p$ and $75$ for $n$.

${\sigma }_{\hat{p}}=\sqrt{\frac{0.15(1-0.15)}{75}}$

$=\sqrt{\frac{0.45\times 0.55}{75}}$

$=\sqrt{0.0033}$

$\approx 0.057445$

Therefore, the standard deviation is $0.057445$.