Question

7. In a clinical trial of the effectiveness of an herb for preventing colds, the results in the accompanying table were obtained. Use a 0.01 significance level to test the claim that getting a cold is independent of the treatment group. What do the results suggest about the effectiveness of the herb as a prevention against colds? 1 Click the icon to view the data on herb treatments Determine the null and alternative hypotheses OA. Ho: The herb treatment does not prevent colds. H1: The herb treatment prevents colds. B. Ho: The herb treatment prevents colds H1: The herb treatment does not prevent colds. Ho: Getting a cold and treatment group are independent. H1: Getting a cold and treatment group are not independent. Ho: Getting a cold and treatment group are not independent. H1: Getting a cold and treatment group are independent. C. ○ D. Determine the test statistic. ,2 Determine the P-value of the test statistic. P-value- What is the conclusion based on the hypothesis test? Since the P-value is (1) of the treatment group. The herb treatment (4) 1: Herb Treatments (Round to three decimal places as needed.) (Round to three decimal places as needed.) the significance level, (2) the null hypothesis. There (3) sufficient evidence to warrant rejection of the claim that getting a cold is independent to be effective as a prevention against colds. Treatment Group Placebo | Herb 20% Extract | Herb 60% Extract Got a Cold Did Not Get a Cold 84 14 54 12 (1) O less than or equal to (2) O fail to reject (3) O is not (4) O does not appear O greater than reject is O appears

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Answer #1

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

\chi2 =4.955

p value =0.084

since the p value is greater than .........2)fail to reject ...........3) is not .............4) does not appear

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