Question

I want red! A candy maker offers Child and Adult bags of jelly beans with different color mixes. The company claims that the Child mix has $30$% red jelly beans while the Adult mix contains $15$% red jelly beans. Assume that the candy maker’s claim is true. Suppose we take a random sample of $50$jelly beans from the Child mix and a separate random sample of $100$jelly beans from the Adult mix.

(a) Find the probability that the proportion of red jelly beans in the Child sample is less than or equal to the proportion of red jelly beans in the Adult sample. Show your work

(b) Suppose that the Child and Adult samples contain an equal proportion of red jelly beans. Based on your result in part (a), would this give you a reason to doubt the company’s claim? Explain.

Short answer

From the given information,

a) The probability that the proportion of red jelly beans in the Child sample is less than or equal to the proportion of red jelly beans in the Adult sample is $0.0212$.

b) Yes, the company claim can be doubted.

Step-by-step solution

Part (a) Step 1: Given Information

It is given in the question that,

The proportion of red jelly beans in the child mix =$30$%

The proportion of red jelly beans in the adult mix =$15$%

Random samples of jelly beans taken from the Child Mix =$50$

Random samples of jelly beans taken from the Adult Mix =$100$

Part(a) Step 2: Explanation

Red jelly beans in bags for children and $p2$is the actual proportion of red jelly beans in bags for adults, the sampling distribution of $p1-p2$is Normal where $p1=\frac{30}{100}=0.30,p2=\frac{15}{100}=0.15$because:

localid="1650446424637" $$\begin{gathered} n_1{p}_1=50\times 0.30 \\ =15 \end{gathered}$$

localid="1650446436131" $$\begin{gathered} n_1(1-p_1)=50\times 0.70 \\ =35 \end{gathered}$$

localid="1650446447320" $$\begin{gathered} n_2{p}_2=100\times 0.15 \\ =15 \end{gathered}$$

localid="1650446461437" $$\begin{gathered} n_2(1-p_2)=100\times 0.85 \\ =85 \end{gathered}$$

$\text{Are at least}10\text{, the sampling distribution of}{\hat{p}}_1-{\hat{p}}_2\text{is approximately Normal. Its mean is:}$

${\mu }_{\dot{\mu }-{\dot{p}}_2}=p_1-p_2$

$=0.30-0.15$

=$0.15$

Part(a) Step 3: Explanation

The standard deviation is :

${\mathit{\sigma }}_{\dot{h}-{\hat{h}}_2}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}$

$=\sqrt{\frac{0.3\times (1-0.3)}{50}+\frac{0.15\times (1-0.15)}{100}}$

$=0.0740$

Hence,

$p({\hat{p}}_1-{\hat{p}}_2\le 0)=p(z\le \frac{0-0.15}{0.0740})$

$=P(z\le -2.03)$

$=0.0212$

Part(b) Step 1: Given Information

It is given in the question, Suppose that the Child and Adult samples contain an equal proportion of red jelly beans. Based on your result in part (a), would this give you a reason to doubt the company’s claim? Explain

Part(b) Step 2: Explanation

Yes, the company's assertion has been questioned. The probability that the proportion of red jelly beans in the Child sample is less than or equal to that in the Adult sample is $0.0212$, based on section (a). If the company's claim is true, there is only a $2$% chance of getting the same number of red jellybeans in the child sample as in the adult sample. This is highly unlikely.