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Name: A type of new car is offered for sale with 4 option packages. A custom
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Description: A type of new car is offered for sale with 4 option packages. A customer can buy any number of these, from none to all 4. A manager proposes the null hypothesis that customers pick packages at random, implying the number of packages bought by a cus

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.8: Probability

Problem 31E

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A type of new car is offered for sale with 4 option packages. A customer can buy any number of these, from none to all 4. A manager proposes the null hypothesis that customers pick packages at random, implying the number of packages bought by a customer should be binomial with

n=4.

This table shows the number of packages chosen by

300

customers.

(a) Find the binomial parameter p needed to calculate the expected counts.

(b) Find the estimated probability that a customer picks one option.

χ2.

(d) Find the value of

χ2

for testing

H0.

(e) Find the p-value for testing

Packages	Customers
0	23
1	80
2	117
3	78
4	2

(a) The binomial parameter p is

(Round to three decimal places as needed.)

Expert Solution

Step 1

Since you have posted a question with multiple sub-parts, we will solve the first three subparts for you. To get the remaining sub-parts solved please repost the complete question and mention the sub-parts to be solved.

a) parameter p is the probability of picking a number of packages at random.

Total no of packages is given to be = 4

So, the sample space of the event of selecting a number of packages by customers will be-

{0, 1, 2, 3, 4}

There are a total of 5 outcomes, so the probability of anyone of the above outcomes occurring is given by $\frac{1}{5} = 0.2$

Hence, p = 0.2

b) Let X be the no of packages bought by customer.

To calculate P[X=1]

Therefore,

$P [X = 1] = (\begin{matrix} 4 \\ 1 \end{matrix}) {(0.2)}^{1} {(1 - 0.2)}^{4 - 1} = (\begin{matrix} 4 \\ 1 \end{matrix}) (0.2) {(0.8)}^{3} = 0.4096$

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