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notsponge [240]
3 years ago
13

A candy maker offers Child and Adult bags of jelly beans with different color mixes. The company claims that the Child mix has 3

0% 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. Let ^ p C and ^ p A be the sample proportions of red jelly beans from the Child and Adult mixes, respectively.
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, assuming that the company’s claim is true.

Enter the probability here:
Mathematics
1 answer:
neonofarm [45]3 years ago
6 0
If the companies claim is to be true this statement would also be true. Since the child bag is halved in the amount of the adults, the percentage would double. There are both 15 red jelly beans in each bag. 15% of 100 is 15. 30% of 50 is 15.
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(a) Find a 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their wate
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Answer:

The margin of error (E)  =0.49

Step-by-step explanation:

Given,

 mean \bar{x}=6.88

 standard deviation \sigma=1.90

 sample size n=41

90\% \quad \text{confidence interval}

Critical value =\frac{z_\alpha}{2}\quad \quad\quad[\because\;\alpha=1-\text{confidence interval}]  

                                                =1-0.90

                                        \Rightarrow \alpha=0.1

Critical value =\frac{z_{0.1}}{2}

                      =z_{0.05}

                      =1.645

The margin of error (E)  =\frac{z_\alpha}{2} \times\frac{\sigma}{\sqrt{n} }

                                       =1.645\times\frac{1.90}{\sqrt{41} }

                                       =0.49

Hence, The margin of error (E) =0.49

Complete question is attached in below.

                     

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