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Verdich [7]
3 years ago
8

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.
Mathematics
1 answer:
malfutka [58]3 years ago
4 0

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 : X^{2}  ∑ ( 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

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fenix001 [56]

Answer:

The lower bound of a 99% C.I for the proportion of defectives = 0.422

Step-by-step explanation:

From the given information:

The point estimate = sample proportion \hat p

\hat p = \dfrac{x}{n}

\hat p = \dfrac{55}{100}

\hat p = 0.55

At Confidence interval of 99%, the level of significance = 1 - 0.99

= 0.01

Z_{\alpha/2} =Z_{0.01/2} \\ \\ = Z_{0.005} = 2.576

Then the margin of error E = Z_{\alpha/2} \times \sqrt{\dfrac{\hat p(1-\hat p)}{n}}

E = 2.576 \times \sqrt{\dfrac{0.55(1-0.55)}{100}}

E = 2.576 \times \sqrt{\dfrac{0.2475}{100}}

E = 2.576 \times0.04975

E = 0.128156

E ≅ 0.128

At 99% C.I for the population proportion p is: \hat p - E

= 0.55 - 0.128

= 0.422

Thus, the lower bound of a 99% C.I for the proportion of defectives = 0.422

6 0
2 years ago
Find each product mentally using the Distributive Property. Show the steps that you used. For 6x13 in distributive property​
aivan3 [116]

Using Distributive Property, the product of 6x13 is 78.

<h3>What is Distributive property?</h3>

The distributive property states that multiplying the sum of two or more addends by a number yields the same outcome as multiplying each addend separately by the number and combining the resulting products.

The steps of distributive property are -

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                                              6 × (10 + 3)

                                          (6 × 10) + (6 × 3)

                                  ((3 + 3) × (5 +5)) + ((3 + 3) × 3)

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                                     (15 + 15 +15 + 15 ) + (9 + 9)

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To learn more about distributive property from given link

brainly.com/question/2807928

#SPJ4

3 0
1 year ago
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3 0
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Step 1: Simplify both sides of the equation.

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8x−40=36

Step 2: Add 40 to both sides.

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Step 3: Divide both sides by 8.

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