applying chi square test:
Observed | OBS | placebo | Herb: 20% extract | Herb:60% extract | Total |
got a cold | 84 | 54 | 44 | 182 | |
not got a cold | 14 | 4 | 12 | 30 | |
Total | 98 | 58 | 56 | 212 | |
Expected | Ei=?row*?column/?total | placebo | Herb: 20% extract | Herb:60% extract | Total |
got a cold | 84.13 | 49.79 | 48.08 | 182 | |
not got a cold | 13.87 | 8.21 | 7.92 | 30 | |
Total | 98 | 58 | 56 | 212 | |
chi square | =(Oi-Ei)2/Ei | placebo | Herb: 20% extract | Herb:60% extract | Total |
got a cold | 0.00 | 0.36 | 0.35 | 0.70 | |
not got a cold | 0.00 | 2.16 | 2.10 | 4.25 | |
Total | 0.00 | 2.51 | 2.44 | 4.955 |
2 =4.955
p value =0.084
since the p value is greater than .........2)fail to reject ...........3) is not .............4) does not appear
7. In a clinical trial of the effectiveness of an herb for preventing colds, the results...
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Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 45 of the 51 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 85 of the 102 subjects developed rhinovirus infections. Use a 0.05 significance level to test the claim that echinacea has an effect on rhinovirus infections. Complete parts (a) through (c) below. a. Test the claim using a hypothesis test. Consider the first sample to be the sample of subjects treated with...
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