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You may need to use the appropriate technology to answer this question. The Wall Street Journal Corporate Perceptions Study 2011 surveyed readers and asked how each rated the quality of management and the reputation of the company for over 250 worldwide corporations. Both the quality of management and the reputation of the company were rated on an excellent, good, and fair categorical scale. Assume the sample data for 200 respondents below applies to this study.

Quality of Management Reputation of Company
Excellent Good Fair
Excellent 40 25 8
Good 35 35 10
Fair 25 10 12

Use a 0.05 level of significance and test for independence of the quality of management and the reputation of the company.

a. State the null and alternative hypotheses.
b. Find the value of the test statistic. (Round your answer to three decimal places.)
c. Find the p-value. (Round your answer to four decimal places.)

Respuesta :

ajeigbeibraheem

Answer:

Step-by-step explanation:

The null and alternative hypothesis:

[tex]H_o \text{: Quality of management and reputation of company are independent}[/tex]

[tex]H_a \text{: Quality of management is not independent of the reputation of company }[/tex]

The observed values:

                      Excellent     Good    Fair    Total

Excellent           40               25        8        73

Good                 35               35        10       80

Fair                     25              10         12       47

Total                 100             70          30      200

The expected values  = (Row total/column total)/grand total)

                      Excellent     Good        Fair    

Excellent           36.5          25.55      10.95  

Good                  40             28             12

Fair                    23.5         16.45         7.05

O                E                    (O - E)²/E

40              36.5                0.3356

35              40                   0.6250

25              23.5                0.0957

25              25.55              0.0118

35              28                    1.7500

10               16.45                2.5290

8                 10.95               0.7947

10                12                    0.3333

12                 7.05                3.4755

                                            9.951

[tex]X^2 = \sum \dfrac{(O-E)^2}{E} \\ \\ \mathbf{X^2 = 9.951}[/tex]

Degree of freedom [tex]df = (r-1)(c-1)[/tex]

[tex]df = (3 - 1 ) (3 - 1)[/tex]

[tex]df = (2)(2)[/tex]

[tex]df = 4[/tex]

Using Excel formula =CHIDIST(x²,d.f)

The P-value = 0.0413

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