91影视

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 =$100$

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" $$\begin{gathered} n_1{p}_1=50脳0.30 \\ =15 \end{gathered}$$

localid="1650446436131" $$\begin{gathered} n_1(1鈭�p_1)=50脳0.70 \\ =35 \end{gathered}$$

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

localid="1650446461437" $$\begin{gathered} n_2(1鈭�p_2)=100脳0.85 \\ =85 \end{gathered}$$

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

${渭}_{\stackrel{藱}{渭}鈭�{\stackrel{藱}{p}}_2}=p_1鈭�p_2$

$=0.30-0.15$

=$0.15$

The standard deviation is :

${\mathit{蟽}}_{\stackrel{藱}{h}鈭�{\hat{h}}_2}=\sqrt{\frac{p_1(1鈭�p_1)}{n_1}+\frac{p_2(1鈭�p_2)}{n_2}}$

$=\sqrt{\frac{0.3脳(1鈭�0.3)}{50}+\frac{0.15脳(1鈭�0.15)}{100}}$

$=0.0740$

Hence,

$p({\hat{p}}_1鈭�{\hat{p}}_2鈮�0)=p(z鈮�\frac{0鈭�0.15}{0.0740})$

$=P(z鈮�鈭�2.03)$

$=0.0212$

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鈥檚 claim? Explain

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.