HL: Chi-Square Tests | Six Minutes: IB Mathematics: Applications and Interpretation, Part 35
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HL: Chi-Square Tests | Six Minutes: IB Mathematics: Applications and Interpretation, Part 35
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1 просмотр · 11 дн. назад
Six Minutes: IB Mathematics: Applications and Interpretation, Part 35: HL: Chi-Square Tests: Stats & Probability — goodness of fit and independence; then HL: t-Tests
Two HL chi-square tests: goodness of fit for one set of categories, and independence for a two-way table. A fair-die example is solved by hand and checked with the GDC chi-square test. Then: How to run one-sample and two-sample t-tests on the GDC to test claims about means. Setting hypotheses, reading the p-value, and concluding in context.
Chapters
0:00 Introduction
0:22 Goodness of fit
0:42 Independence
1:09 Worked example
2:22 Trap
2:39 Recap
3:03 HL: t-Tests
3:18 What a t-test does
3:49 One-sample t-test
4:14 Two-sample t-test
4:44 Worked example
5:56 Trap
6:09 Recap
6:26 Thanks for watching
Practice problems
1. A spinner has 4 equal-looking sectors. In 80 spins the counts are 12, 26, 19, 23. Test at the 5% level whether all sectors are equally likely. (E = 20 each; χ² = (64+36+1+9)/20 = 5.5; df = 3; p ≈ 0.139 > 0.05, so do not reject H0 — no evidence the spinner is biased.)
2. 100 students: 40 juniors (30 prefer text, 10 prefer video) and 60 seniors (28 prefer text, 32 prefer video). Test at the 5% level whether preference is independent of year. (Expected: juniors text 23.2, video 16.8; seniors text 34.8, video 25.2; χ² ≈ 7.91; df = 1; p ≈ 0.005 < 0.05, so reject H0 — preference is not independent of year.)
3. A factory claims its bolts average 25 mm. A sample of 16 bolts has mean 24.7 mm and sample standard deviation 0.9 mm. Test at the 5% level whether the mean differs from 25 (two-tailed). t ≈ −1.33, df = 15, p ≈ 0.20 > 0.05, so fail to reject H0: no evidence the mean differs from 25 mm.
4. Five plants given fertilizer A grow 12.1, 12.5, 11.8, 12.3, 12.0 cm; five given B grow 12.8, 13.1, 12.6, 13.0, 12.9 cm. Test at 5% whether B's mean is greater than A's. Two-sample t-test gives t ≈ −4.99, df = 8 (pooled), p ≈ 0.0005 < 0.05, so reject H0: B's mean is greater.
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