54 lines
1.8 KiB
Markdown
54 lines
1.8 KiB
Markdown
# Homework 10 - Aidan Sharpe
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## 1
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The titanium content in an aircraft-grade alloy is an important discriminant of strength. A sample of 10 test coupns reveals the following titanium content in percent:
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| Coupon | Titanium Content |
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| ------ | ---------------- |
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| 1 | 8.30% |
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| 2 | 8.09% |
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| 3 | 8.99% |
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| 4 | 8.60% |
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| 5 | 8.40% |
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| 6 | 8.35% |
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| 7 | 8.36% |
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| 8 | 8.75% |
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| 9 | 8.91% |
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| 10 | 8.05% |
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Suppose that the distribution of titanium content is symmetric and continuous. Does the sample data suggest that the mean titanium content differs significantly from 8.5%? Use $\alpha = 0.05$
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```python
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>>> coupons = [8.3, 8.09, 8.99, 8.60, 8.40, 8.35, 8.36, 8.75, 8.91, 8.05]
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>>> mu0 = 8.5
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# Find the difference between each sample and mu0
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>>> differences = [xi - mu0 for xi in coupons]
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# Find the absolute differences
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>>> abs_diffs = [abs(x) for x in differences]
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# Sort the differences in ascending order
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>>> s_diffs = sorted(abs_diffs)
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# Turn the sorted order into a ranked list
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>>> ranks = [abs_diffs.index(x) + 1 for x in s_diffs]
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# Find the ranks corresponding to positive differences
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>>> p_ranks = [ranks[i] if differences[i] > 0 else 0 for i in range(len(differences))]
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# Find the ranks corresponding to negative differences
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>>> n_ranks = [ranks[i] if differences[i] < 0 else 0 for i in range(len(differences))]
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# Find the positive and negative rank sums
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>>> wp = sum(p_ranks)
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>>> wn = sum(n_ranks)
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# Find the test statistic
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>>> w_observed = min(wp, wn)
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>>> w_observed
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22
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```
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For a two-sided signed rank test with $\alpha = 0.05$ and 10 samples, $w_\alpha^*$ is 8. Since 22 is greater than 8, we do not have enough evidence to suggest that the mean titanium content differs from 8.5%.
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