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Chapter 11: Q.9 (page 654)

A teacher predict that the distribution of grades on the final exam will be and they are recorded in table 11.27

The actual distribution for a class of 20 is in table 11.28

df=----

Short Answer

Expert verified

df=3

Step by step solution

01

Given Information

Given tables are

we have to determine the degree of freedom.

02

Explanation

The formula of degree of freedom is

df=n-1

where nis number of cells

df=n-1=4-1=3

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Most popular questions from this chapter

A 2013poll in California surveyed people about taxing sugar-sweetened beverages. The results are presented in the Table and are classified by ethnic group and response type. Are the poll responses independent of the participants’ ethnic group? Conduct a test of independence at the 5%significance level.

OpinionlEthnicityAsian-AmericanWhite/Non-HispanicAfrican-AmericanLatinoRowTotalAgainst tax4843341160682In Favor of tax5423424147459No opinion1643161994Column Total118710813261235

A teacher predict that the distribution of grades on the final exam will be and they are recorded in table 11.27

The actual distribution for a class of 20 is in table 11.28

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The expected percentage of the number of pets students have in their homes is distributed (this is the given distribution for the student population of the United States) as in Table 11.12.

Number of PetsPercent0181252303184+9

A random sample of 1,000students from the Eastern United States resulted in the data in Table 11.13.

Number of PetsFrequency02101240232031404+90

At the 1% significance level, does it appear that the distribution “number of pets” of students in the Eastern United States is different from the distribution for the United States student population as a whole? What is the p-value?

Read the statement and decide whether it is true or false.

In a goodness of fit test, if the p-value is 0.0113, in general, do not reject the null hypothesis

The manager of "Frenchies" is concerned that patrons are not consistently receiving the same amount of French fries with each order. The chef claims that the standard deviation for a ten-ounce order of fries is at most 1.5oz., but the manager thinks that it may be higher. He randomly weighs49 orders of fries, which yields a mean of 11 oz. and a standard deviation of two oz.
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