Question

In: Statistics and Probability

Groups of 3 Person 1 Choose 10 5-letter words without knowledge of Person 3 Choose 10...

Groups of 3

Person 1

Choose 10 5-letter words without knowledge of Person 3

Choose 10 7 letter words without knowledge of Person 3

Person 2

Scramble the letters of the words

Timer

Person 3

Unscramble the words

Data

Word Number

Letters

1

2

3

4

5

6

7

8

9

10

Total

5

2.28

5.36

10.48

5.92

3.32

4.49

4.29

4.31

4.07

5.85

032.25

7

4.02

7.55

15.16

6.76

7.22

13.61

42.86

7.03

5.95

7.07

117.23

Conduct a test of hypothesis to determine if your person #3 has a lower
average time to unscramble 5-letter words than 7-letter words.

Note... this question was asked previously but with out the instruction... when it came back to me, there was no way to edit. So I'm hoping not to be charged a question in this venue. Thanks.

Solutions

Expert Solution

Data:      

n1 = 10     

n2 = 10     

x1-bar = 5.037     

x2-bar = 11.723     

s1 = 2.2121     

s2 = 11.4676     

Hypotheses:      

Ho: μ1 ≥ μ2      

Ha: μ1 < μ2      

Decision Rule:      

α = 0.05     

Degrees of freedom =   10 + 10 - 2 = 18   

Critical t- score =   -1.734063592    

Reject Ho if t <   -1.734063592    

Test Statistic:      

Pooled SD, s = √[{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)] =      √(((10 - 1) * 2.2121^2 + (10 - 1) * 11.4676^2) / (10 + 10 - 2)) = 8.258306006

SE = s * √{(1 /n1) + (1 /n2)} = 8.25830600577382 * √((1/10) + (1/10)) = 3.693226722    

t = (x1-bar -x2-bar)/SE = (5.037 - 11.723)/3.69322672158101 = -1.810341066    

p- value = 0.043484726     

Decision (in terms of the hypotheses):      

Since -1.810341066 < -1.734063592 we reject Ho and accept Ha

Conclusion (in terms of the problem):

There is sufficient evidence that this person takes significantly less time to unsramble 5-letter words than 7-letter words.


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