Question

The candy machine Suppose a large candy machine has $15\%$orange candies. Imagine taking an localid="1652783380833" $SRS$of localid="1652783389017" $25$candies from the machine and observing the sample proportion localid="1652783396603" $\hat{p}$of orange candies.

(a) What is the mean of the sampling distribution of localid="1652783403494" $\hat{p}$? Why?

(b) Find the standard deviation of the sampling distribution of localid="1652783417404" $\hat{p}$. Check to see if the localid="1652783410131" $10\%$condition is met.

(c) Is the sampling distribution of localid="1652783430137" $\hat{p}$approximately Normal? Check to see if the Normal condition is met.

(d) If the sample size were localid="1652783436404" $75$rather than localid="1652783443101" $25$, how would this change the sampling distribution of localid="1652783450320" $\hat{p}$

Short answer

a). The required mean is $0.15$.

b). The required standard deviation is localid="1652784804305" $0.0714143$.

c). localid="1652784834600" $\hat{p}$sample distribution is not close to Normal.

d). The standard deviation islocalid="1652784839897" $0.57445$.

Step-by-step solution

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 $0.057445$.