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 mangement and the reputation of the company were rated on an excellent, good, and fair catergorical scale. Assume sample data for 200 respondents below. Reputation of Company
Quality of management Excellent Good Fair
Excellent 40 25 5
Good 35 35 10
Fair 25 10 15
A) use a .05 level of significance and test for independence of the quality of management and the reputation of the company. What is the p-value and what is your conculsion??
B) If there is a dependence or association between the two ratings, discuss and use the probabilities to justify your answer.

Hello some parts of your question is missing below is the missing part

answer : A) p-value =  0.0019 ,

The level of significance ( 0.05 ) > p-value ( 0.0019 )

B) There is no dependence between the two ratings

Step-by-step explanation:

A) using a 0.05 level of significance and test for independence of the quality of management

Test statistic :  ∑ ( Oi - Ei )^2 / Ei

= [(40-35)^2 / 35 ] +  [(25-24.5)^2 / 24.5 ] +  [(5-10.5)^2 / 10.5]  +  [(35-40)^2 / 40]  +  [(35-28)^2 / 28]  +  [(10-12)^2 / 12]  +   [ (25-25)^2 / 25 ]  +  [(10-17.5)^2 / 17.5 ]  + [(15-7.5)^2 / 7.5]  =  17.03  

p-value = 0.0019

Df = 4  ,

The level of significance ( 0.05 ) > p-value ( 0.0019 )

B) There is no dependence between the two ratings

h df tyjgtyj ttd y ytkgfty ty

step-by-step explanation:

tyty jry j y yj yttj yhgtyj y ghyj ytthj

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}