Question Help e below shows the overal scores and the coot per load for eight di

Question

Question Help e below shows the overal scores and the coot per load for eight dishwasher detergents The overal score represents the cleaning test results for the he overall skore and the price? Complete parts a through e below the cleaning test results for the d h washer deler ent At -0 01, can you conclude that there is a come ation between Overall scovre 68 3 7987 73 64 70 74e score 68 9 79 87 73 4 7074 in cents Cick the 'con In view te critical values for the Spearman rank comelation (a) identify the claim and state Ho and Ha Ho There is H, There is The | between the averall scare and the price. | between the overall score and the price. Is the claim (b) Determine the critical value. The critical value is Round to tree decimal places as needed) (c) Find the Spearman rank correlation coeficient Round to three decimal places as needed) (d) Decide whether to reject or fail to reject the null hypothesis O A. Since i, is greater than the critical value, reject Ho O8. Since Is, is greater than the critical value, fail to reject Ho O C, Since It is less than the critical value fail to reject Ho Click to select your answer(s) 7:35 PM 79/2017 2L

Explanation / Answer

Part (a)

Claim: There is no correlation between x and y.[i.e., Spearman Rank Correlation Coefficient iszero.]

Part (b)

Given sample size is 8 and level of significance as 0.01, Critical value is the upper 0.5% [half of 0.01] of t6 i.e., t-distribution with 6 (i.e., n - 2) degrees of freedom.

Using Excel Function, the above value is found to be 3.707 ANSWER

Part (c)

Value of Spearman Rank Correlation Coefficient, rxy = 0.530 ANSWER

Formula for rxy = 1 – [{6[1,n](di2)}/{{n(n2 - 1)}] where

n = sample size, di = Rank(xi) – Rank(yi).

Computation details are given below:

i

xi

yi

Rx

Ry

di =Rx - Ry

di^2

1

17

68

6

2

4

16

2

14

79

4.5

6.5

-2

4

3

24

79

8

6.5

1.5

2.25

4

19

87

7

8

-1

1

5

5

73

1

4

-3

9

6

10

64

2

1

1

1

7

14

70

4.5

3

1.5

2.25

8

13

74

3

5

-2

4

n

8

sum(di^2)

39.5

alpha

0.01

6sumdi^2

237

n(n^2 - 1)

504

r

0.52976

r^2

0.28064

1 - r^2

0.71935

n - 2

6

(n-2)/(1-r^2)

8.34083

sqrt(n-2)/(1-r^2)

2.88805

t

1.52997

tcrit

3.70742

Part (d)

Decision

Since calculated value of t (1.530) < the critical value, the null hypothesis that correlation is zero is accepted.

Hence, the claim is valid at 1% level of significance. ANSWER

i

xi

yi

Rx

Ry

di =Rx - Ry

di^2

1

17

68

6

2

4

16

2

14

79

4.5

6.5

-2

4

3

24

79

8

6.5

1.5

2.25

4

19

87

7

8

-1

1

5

5

73

1

4

-3

9

6

10

64

2

1

1

1

7

14

70

4.5

3

1.5

2.25

8

13

74

3

5

-2

4

n

8

sum(di^2)

39.5

alpha

0.01

6sumdi^2

237

n(n^2 - 1)

504

r

0.52976

r^2

0.28064

1 - r^2

0.71935

n - 2

6

(n-2)/(1-r^2)

8.34083

sqrt(n-2)/(1-r^2)

2.88805

t

1.52997

tcrit

3.70742

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