The herb ginkgo biloba is commonly used as a treatment to prevent dementia. In a

Question

The herb ginkgo biloba is commonly used as a treatment to prevent dementia. In a study of the effectiveness of this treatment, 1545 elderly subjects were given ginkgo and 1524 elderly subjects were given a placebo. Among those in the ginkgo treatment group, 246 later developed dementia, and among those in the placebo group, 277 later developed dementia (based on data from "Ginkgo Biloba for Prevention of Dementia, " by Demoski et al., Journal of American Medical Association, Vol. 300, No. 19). We want to use a 0.01 significance level to test the claim that ginkgo is effective in preventing dementia. Identify the test statistic, p -value or critical value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. z = -1.66, critical value: z = -1.96. P- value 0.78, reject H_0 z = 7.21, critical value: z = 2.37, P- value 0.095, fail to reject H_0 Test statistic: z = 5412. critical value: z = 1.34. P-value 0.23, reject H_0 Test statistic: z = 7.23, critical value: z = 1.82. P- value 0.0001, fail to reject H_0 Test statistic: z = -1.66. critical value: z = -2.33, p-value 0.0485, fail to reject H_0

Explanation / Answer

Using Minitab.

Test and CI for Two Proportions

Sample X N Sample p
1 246 1545 0.159223
2 277 1524 0.181759

Difference = p (1) - p (2)
Estimate for difference: -0.0225352
95% upper bound for difference: -0.000209111
Test for difference = 0 (vs < 0): Z = -1.66 P-Value = 0.048

Here P-value 0.048 > 0.01 we fail to reject H0

Here Z=-1.66 p-value =0048 and critical z va;lue =-2.33

so Option 5th is the correct choice

Hope this will be helpful to you. Thanks :)

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