MATH SOLVE

6 months ago

Q:
# It has been observed that some persons who suffer acute heartburn, again suffer acute heartburn within one year of the first episode. This is due, in part, to damage from the first episode. The performance of a new drug designed to prevent a second episode is to be tested for its effectiveness in preventing a second episode. In order to do this two groups of people suffering a first episode are selected. There are 55 people in the first group and this group will be administered the new drug. There are 45 people in the second group and this group will be administered a placebo. After one year, 11% of the first group has a second episode and 9% of the second group has a second episode. Conduct a hypothesis test to determine, at the significance level 0.1, whether there is reason to believe that the true percentage of those in the first group who suffer a second episode is different from the true percentage of those in the second group who suffer a second episode? Select the [Rejection Region, Decision to Reject (RH0) or Failure to Reject (FRH0)] A. [ z < -1.65, RHo] B. [ z < -1.65 and z > 1.65, FRHo C. [z > 1.65, FRHo] D. [z < -1.65 and z > 1.65, FRHo] E. [z > -1.65 and z < 1.65, RHo] F. None of the above

Accepted Solution

A:

Using the z-distribution, it is found that since the absolute value of the test statistic is less than the absolute value of the critical value for the two-tailed test, there is no reason to believe that the true percentage of those in the first group who suffer a second episode is different from the true percentage of those in the second group who suffer a second episode.At the null hypothesis, it is tested if the proportions are not different, that is, their subtraction is 0, hence:[tex]H_0: p_1 - p_2 = 0[/tex]At the alternative hypothesis, it is tested if they are different, that is, their subtraction is not 0, hence:[tex]H_1: p_1 - p_2 \neq 0[/tex]For each sample, the proportion and the standard error are:[tex]p_1 = 0.13, s_1 = \sqrt{\frac{0.13(0.87)}{163}} = 0.0263[/tex][tex]p_2 = 0.14, s_2 = \sqrt{\frac{0.14(0.86)}{160}} = 0.0274[/tex]The mean and the standard error of the distribution of the difference is given by:[tex]\overline{p} = p_2 - p_1 = 0.14 - 0.13 = 0.01[/tex][tex]s = \sqrt{s_1^2 + s_2^2} = \sqrt{0.0263^2 + 0.0274^2} = 0.038[/tex]The test statistic is:[tex]z = \frac{\overline{p} - p}{s}[/tex]In which [tex]p = 0[/tex] is the value tested at the null hypothesis.Hence:[tex]z = \frac{0.01}{0.038}[/tex][tex]z = 0.26[/tex]The critical value for a two-tailed test, as we are testing if the mean is different of a value, with a significance level of 0.1 is [tex]z^{\ast} = \pm 1.645[/tex]Since the absolute value of the test statistic is less than the absolute value of the critical value for the two-tailed test, there is no reason to believe that the true percentage of those in the first group who suffer a second episode is different from the true percentage of those in the second group who suffer a second episode.A similar problem is given at