Flaked | Nonflaked | Difference |
8 | 3 | 5 |
8 | 1 | 7 |
10 | 2 | 8 |
1 | 3 | -2 |
4 | 7 | -3 |
38 | 32 | 6 |
51 | 30 | 21 |
25 | 12 | 13 |
Sample mean of the difference using excel function AVERAGE(), x̅d = 6.8750
Sample standard deviation of the difference using excel function STDEV.S(), sd = 7.7356
Sample size, n = 8
a) Level of significance = 0.05
Null and Alternative hypothesis:
Ho : µd = 0
H1 : µd > 0
It is right tailed.
b) The student's t. We assume that d has an approximately normal distribution.
Test statistic:
t = (x̅d)/(sd/√n) = (6.875)/(7.7356/√8) = 2.5138
df = n-1 = 7
c) p-value :
Right tailed p-value = T.DIST.RT(2.5138, 7) = 0.0201
0.005 < p-value < 0.025
d) At the α = 0.05 level, we reject the null hypothesis and conclude that data are statistically significant.
e) Reject the null hypothesis. There is sufficient evidence to claim that the average number of flaked stone tools is higher.
of the studen t the cost of that is in some stations, the choice of the...
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