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Chapter 10: Q 1. (page 639)

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 50jelly beans from the Child mix and a separate random sample of 100jelly beans from the Adult mix. Let p^Cand p^Abe the sample proportions of red jelly beans from the Child and

Adult mixes, respectively.

a. What is the shape of the sampling distribution of p^C-p^A? Why?

b. Find the mean of the sampling distribution.

c. Calculate and interpret the standard deviation of the sampling distribution.

Short Answer

Expert verified

a. The shape in normal.

b. μp^C-p^A=0.15

c.σp^C-p^A=0.07399

Step by step solution

01

Given Information

It is given that nC=50

nA=100

pC=0.30

pA=0.15

02

Shape of p^C-p^A

Assuming p^C-p^Ais normal.

Condition is:

nCpC≥10

nC1-pC≥10

nApA≥10

nCpC=(50)(0.30)=15

nApA=(100)(0.15)=15

nA1-pA=(100)(1-0.15)=(100)(0.85)=85

The shape is approximately normal as it satisfies all the four conditions.

03

Mean of Sampling Distribution

Using μp^C-p^A=pC-pA

=0.30-0.15=0.15

Mean is0.17

04

Standard Deviation

Child jelly(50)<10%of all child jelly beansand

Adult jelly(100)<10%of all adult jelly beans

Standard deviation is σp^C-p^A=pC1-pCnC+pA1-pAnA

=0.30(1-0.30)50+0.15(1-0.15)100

=0.30(0.70)50+0.15(0.85)100≈0.07399

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