You are conducting a goodness of fit hypothesis test (� = 0.01) for the claim that there is a difference in sales for 5 shelves at a grocery store. The number of items sold per shelf are observed and recorded over a week, the results are in the table below. Round answers to 4 decimal places. Retain and use unrounded numbers for future calculations.a. Complete the rest of the table. Shelf LabelObserved FrequencyExpected Frequency(�-�)2� A6 B25 C25 D15 E12 b. Calculate the chi-square test-statistic for this data. �2 = What are the degrees of freedom for this test? d.f. = Compute the p-value for this sample. p-value = The p-value is...less than (or equal to) �greater than �c. This test statistic leads to a decision to...reject the nullaccept the nullfail to reject the nullaccept the alternative As such, the final conclusion is that...There is not sufficient evidence to support the claim that there is a difference in sales for 5 shelves at a grocery store.There is sufficient evidence to warrant rejection of the claim that there is a difference in sales for 5 shelves at a grocery store.The sample data support the claim that there is a difference in sales for 5 shelves at a grocery store.There is sufficient evidence to support the claim that there is a difference in sales for 5 shelves at a grocery store.

Answered: You are conducting a goodness of fit…

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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You are conducting a goodness of fit hypothesis test (� = 0.01) for the claim that there is a difference in sales for 5 shelves at a grocery store. The number of items sold per shelf are observed and recorded over a week, the results are in the table below. Round answers to 4 decimal places. Retain and use unrounded numbers for future calculations.

a. Complete the rest of the table.

Shelf Label	Observed Frequency	Expected Frequency	(�-�)2�
A	6
B	25
C	25
D	15
E	12

b. Calculate the chi-square test-statistic for this data.
�2 =

What are the degrees of freedom for this test?
d.f. =

Compute the p-value for this sample.
p-value =

The p-value is...

less than (or equal to) �
greater than �

c. This test statistic leads to a decision to...

reject the null
accept the null
fail to reject the null
accept the alternative

As such, the final conclusion is that...

There is not sufficient evidence to support the claim that there is a difference in sales for 5 shelves at a grocery store.
There is sufficient evidence to warrant rejection of the claim that there is a difference in sales for 5 shelves at a grocery store.
The sample data support the claim that there is a difference in sales for 5 shelves at a grocery store.
There is sufficient evidence to support the claim that there is a difference in sales for 5 shelves at a grocery store.

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