In: Statistics and Probability
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.
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.