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A. What is the p-value? more than 0.05 less than 0.005 between 0.025 and 0.05 between 0.005 and 0.01 between 0.01 and 0.025 B. Givenn the siggnificance level, a = 5%, whatt’s your final connclusion? Would you connclude thatt there is preferennce for one or morre brands? Fail to reject the null. There is no brand preference. Reject the null. There is no brand preference. Reject the null. There is a preference for one or more brands. Fail to reject the null. There is a preference for one or more brands

A. What is the p-value? more than 0.05 less than 0.005 between 0.025 and 0.05 between 0.005 and 0.01 between 0.01 and 0.025 B. Givenn the siggnificance level, a = 5%, whatt’s your final connclusion? Would you connclude thatt there is preferennce for one or morre brands? Fail to reject the null. There is no brand preference. Reject the null. There is no brand preference. Reject the null. There is a preference for one or more brands. Fail to reject the null. There is a preference for one or more brands

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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A. What is the p-value?

more than 0.05
less than 0.005
between 0.025 and 0.05
between 0.005 and 0.01
between 0.01 and 0.025
B. Givenn the siggnificance level, a = 5%, whatt’s your final connclusion? Would you connclude thatt there is preferennce for one or morre brands?

Fail to reject the null. There is no brand preference.
Reject the null. There is no brand preference.
Reject the null. There is a preference for one or more brands.
Fail to reject the null. There is a preference for one or more brands.
Expected Frequency |(fe) Observed Frequency Brand (fo) (fo - fe)2 / fe |1 78 2 117 3 105 Test statistic: Total 300
Transcribed Image Text:Expected Frequency |(fe) Observed Frequency Brand (fo) (fo - fe)2 / fe |1 78 2 117 3 105 Test statistic: Total 300
Expert Solution
Step 1

Given,

A table of observed frequencies of 3 brands,

 

To test is any brand prefer over other,

Under the null hypothesis, we assume there is no brand preference.

So,

The observed frequencies of each brand are 3003 = 100

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